When Do You Know That You Need to Do the Cahin Rule for Calc

Formula for the derivative of equanimous functions

In calculus, the concatenation rule is a formula that expresses the derivative of the limerick of two differentiable functions f and g in terms of the derivatives of f and g. More precisely, if ${\displaystyle h=f\circ g}$ is the part such that ${\displaystyle h(x)=f(g(x))}$ for every 10, then the chain rule is, in Lagrange's notation,

${\displaystyle h'(x)=f'(g(10))1000'(10).}$

or, equivalently,

${\displaystyle h'=(f\circ g)'=(f'\circ g)\cdot one thousand'.}$

The chain rule may also be expressed in Leibniz's notation. If a variable z depends on the variable y, which itself depends on the variable ten (that is, y and z are dependent variables), then z depends on ten besides, via the intermediate variable y. In this case, the chain rule is expressed as

${\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}},}$

and

${\displaystyle \left.{\frac {dz}{dx}}\right|_{x}=\left.{\frac {dz}{dy}}\right|_{y(x)}\cdot \left.{\frac {dy}{dx}}\right|_{10},}$

for indicating at which points the derivatives have to be evaluated.

In integration, the counterpart to the concatenation dominion is the substitution rule.

Intuitive explanation [edit]

Intuitively, the chain rule states that knowing the instantaneous rate of modify of z relative to y and that of y relative to x allows one to calculate the instantaneous charge per unit of alter of z relative to ten as the product of the two rates of change.

Every bit put by George F. Simmons: "if a automobile travels twice as fast as a cycle and the bike is iv times as fast equally a walking man, and so the car travels 2 × 4 = viii times as fast every bit the human being."^[1]

The relationship betwixt this case and the chain rule is as follows. Let z, y and x be the (variable) positions of the car, the bike, and the walking man, respectively. The rate of change of relative positions of the car and the bicycle is ${\textstyle {\frac {dz}{dy}}=two.}$ Similarly, ${\textstyle {\frac {dy}{dx}}=4.}$ So, the rate of change of the relative positions of the car and the walking homo is

${\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}=ii\cdot 4=8.}$

The rate of modify of positions is the ratio of the speeds, and the speed is the derivative of the position with respect to the time; that is,

${\displaystyle {\frac {dz}{dx}}={\frac {\frac {dz}{dt}}{\frac {dx}{dt}}},}$

or, equivalently,

${\displaystyle {\frac {dz}{dt}}={\frac {dz}{dx}}\cdot {\frac {dx}{dt}},}$

which is as well an application of the chain rule.

History [edit]

The chain rule seems to have first been used by Gottfried Wilhelm Leibniz. He used it to calculate the derivative of ${\displaystyle {\sqrt {a+bz+cz^{ii}}}}$ as the composite of the foursquare root office and the function ${\displaystyle a+bz+cz^{2}\!}$ . He start mentioned information technology in a 1676 memoir (with a sign mistake in the adding). The common notation of the chain rule is due to Leibniz.^[2] Guillaume de l'Hôpital used the concatenation dominion implicitly in his Analyse des infiniment petits. The concatenation dominion does non appear in any of Leonhard Euler'south analysis books, fifty-fifty though they were written over a hundred years after Leibniz'due south discovery.

Statement [edit]

The simplest class of the chain rule is for real-valued functions of 1 real variable. It states that if g is a function that is differentiable at a signal c (i.due east. the derivative k′(c) exists) and f is a role that is differentiable at g(c), then the blended function ${\displaystyle f\circ yard}$ is differentiable at c , and the derivative is^[3]

${\displaystyle (f\circ g)'(c)=f'(chiliad(c))\cdot g'(c).}$

The dominion is sometimes abbreviated as

${\displaystyle (f\circ thou)'=(f'\circ thousand)\cdot m'.}$

If y = f(u) and u = thousand(10), then this abbreviated form is written in Leibniz notation as:

${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dx}}.}$

The points where the derivatives are evaluated may likewise be stated explicitly:

${\displaystyle \left.{\frac {dy}{dx}}\right|_{x=c}=\left.{\frac {dy}{du}}\right|_{u=g(c)}\cdot \left.{\frac {du}{dx}}\correct|_{ten=c}.}$

Carrying the same reasoning further, given due north functions ${\displaystyle f_{1},\ldots ,f_{northward}\!}$ with the composite function ${\displaystyle f_{i}\circ (f_{2}\circ \cdots (f_{north-1}\circ f_{due north}))\!}$ , if each role ${\displaystyle f_{i}\!}$ is differentiable at its immediate input, and so the blended role is also differentiable by the repeated awarding of Chain Rule, where the derivative is (in Leibniz's notation):

${\displaystyle {\frac {df_{1}}{dx}}={\frac {df_{1}}{df_{2}}}{\frac {df_{two}}{df_{3}}}\cdots {\frac {df_{n}}{dx}}.}$

Applications [edit]

Composites of more than two functions [edit]

The chain rule tin be practical to composites of more than than ii functions. To take the derivative of a composite of more than two functions, notice that the composite of f, one thousand, and h (in that order) is the composite of f with thousand ∘ h . The concatenation rule states that to compute the derivative of f ∘ g ∘ h , it is sufficient to compute the derivative of f and the derivative of yard ∘ h . The derivative of f can be calculated directly, and the derivative of g ∘ h can be calculated past applying the chain rule again.

For concreteness, consider the role

${\displaystyle y=e^{\sin(x^{2})}.}$

This tin can be decomposed as the composite of three functions:

${\displaystyle {\brainstorm{aligned}y&=f(u)=e^{u},\\[6pt]u&=k(v)=\sin v=\sin(x^{two}),\\[6pt]5&=h(10)=x^{2}.\end{aligned}}}$

Their derivatives are:

${\displaystyle {\brainstorm{aligned}{\frac {dy}{du}}&=f'(u)=due east^{u}=due east^{\sin(x^{2})},\\[6pt]{\frac {du}{dv}}&=grand'(5)=\cos v=\cos(ten^{2}),\\[6pt]{\frac {dv}{dx}}&=h'(x)=2x.\end{aligned}}}$

The chain rule states that the derivative of their composite at the indicate x = a is:

${\displaystyle {\begin{aligned}(f\circ g\circ h)'(a)&=f'((1000\circ h)(a))\cdot (g\circ h)'(a)\\[10pt]&=f'((m\circ h)(a))\cdot g'(h(a))\cdot h'(a)=(f'\circ g\circ h)(a)\cdot (g'\circ h)(a)\cdot h'(a).\end{aligned}}}$

In Leibniz notation, this is:

${\displaystyle {\frac {dy}{dx}}=\left.{\frac {dy}{du}}\correct|_{u=g(h(a))}\cdot \left.{\frac {du}{dv}}\right|_{v=h(a)}\cdot \left.{\frac {dv}{dx}}\right|_{x=a},}$

or for short,

${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}\cdot {\frac {du}{dv}}\cdot {\frac {dv}{dx}}.}$

The derivative role is therefore:

${\displaystyle {\frac {dy}{dx}}=east^{\sin(ten^{ii})}\cdot \cos(x^{2})\cdot 2x.}$

Another way of computing this derivative is to view the composite function f ∘ g ∘ h equally the composite of f ∘ thousand and h. Applying the chain dominion in this fashion would yield:

${\displaystyle (f\circ g\circ h)'(a)=(f\circ chiliad)'(h(a))\cdot h'(a)=f'(thou(h(a)))\cdot g'(h(a))\cdot h'(a).}$

This is the aforementioned as what was computed to a higher place. This should be expected because (f ∘ thou) ∘ h = f ∘ (k ∘ h).

Sometimes, information technology is necessary to differentiate an arbitrarily long composition of the form ${\displaystyle f_{1}\circ f_{2}\circ \cdots \circ f_{n-1}\circ f_{n}\!}$ . In this case, define

${\displaystyle f_{a\,.\,.\,b}=f_{a}\circ f_{a+1}\circ \cdots \circ f_{b-1}\circ f_{b}}$

where ${\displaystyle f_{a\,.\,.\,a}=f_{a}}$ and ${\displaystyle f_{a\,.\,.\,b}(ten)=x}$ when ${\displaystyle b<a}$ . Then the chain rule takes the form

${\displaystyle Df_{ane\,.\,.\,n}=(Df_{1}\circ f_{2\,.\,.\,n})(Df_{2}\circ f_{3\,.\,.\,north})\cdots (Df_{due north-1}\circ f_{n\,.\,.\,n})Df_{n}=\prod _{g=ane}^{northward}\left[Df_{k}\circ f_{(k+1)\,.\,.\,n}\right]}$

or, in the Lagrange annotation,

${\displaystyle f_{one\,.\,.\,northward}'(x)=f_{1}'\left(f_{2\,.\,.\,n}(x)\right)\;f_{ii}'\left(f_{3\,.\,.\,n}(10)\right)\cdots f_{north-1}'\left(f_{northward\,.\,.\,due north}(x)\right)\;f_{n}'(x)=\prod _{k=1}^{north}f_{k}'\left(f_{(k+1\,.\,.\,northward)}(x)\right)}$

Quotient rule [edit]

The chain dominion can exist used to derive some well-known differentiation rules. For instance, the caliber rule is a issue of the concatenation rule and the product rule. To run into this, write the office f(x)/one thousand(x) every bit the production f(x) · i/g(x). First apply the product rule:

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\left({\frac {f(x)}{g(10)}}\right)&={\frac {d}{dx}}\left(f(10)\cdot {\frac {1}{one thousand(x)}}\right)\\&=f'(ten)\cdot {\frac {1}{grand(x)}}+f(x)\cdot {\frac {d}{dx}}\left({\frac {i}{one thousand(10)}}\right).\end{aligned}}}$

To compute the derivative of 1/g(x), notice that it is the composite of k with the reciprocal part, that is, the function that sends ten to ane/x . The derivative of the reciprocal function is ${\displaystyle -one/ten^{2}\!}$ . By applying the concatenation rule, the concluding expression becomes:

${\displaystyle f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot \left(-{\frac {one}{g(x)^{ii}}}\cdot k'(x)\right)={\frac {f'(x)1000(x)-f(x)chiliad'(x)}{g(x)^{2}}},}$

which is the usual formula for the quotient dominion.

Derivatives of inverse functions [edit]

Suppose that y = chiliad(10) has an inverse function. Telephone call its inverse function f so that we have 10 = f(y). There is a formula for the derivative of f in terms of the derivative of yard. To see this, note that f and yard satisfy the formula

${\displaystyle f(1000(10))=10.}$

And because the functions ${\displaystyle f(g(10))}$ and x are equal, their derivatives must be equal. The derivative of x is the constant role with value ane, and the derivative of ${\displaystyle f(chiliad(10))}$ is adamant by the chain rule. Therefore, we have that:

${\displaystyle f'(m(x))g'(10)=1.}$

To express f' as a function of an contained variable y, we substitute ${\displaystyle f(y)}$ for x wherever it appears. Then nosotros tin solve for f'.

${\displaystyle {\begin{aligned}f'(yard(f(y)))chiliad'(f(y))&=one\\[5pt]f'(y)one thousand'(f(y))&=1\\[5pt]f'(y)={\frac {one}{g'(f(y))}}.\end{aligned}}}$

For instance, consider the function g(ten) = eastward ^ten . It has an changed f(y) = ln y . Because one thousand′(x) = e ¹⁰ , the above formula says that

${\displaystyle {\frac {d}{dy}}\ln y={\frac {i}{east^{\ln y}}}={\frac {ane}{y}}.}$

This formula is true whenever k is differentiable and its inverse f is also differentiable. This formula can neglect when one of these conditions is not true. For example, consider grand(ten) = x ⁱⁱⁱ . Its changed is f(y) = y ^1/3 , which is not differentiable at zero. If nosotros try to use the above formula to compute the derivative of f at cipher, then we must evaluate 1/g′(f(0)). Since f(0) = 0 and k′(0) = 0, we must evaluate 1/0, which is undefined. Therefore, the formula fails in this case. This is not surprising because f is non differentiable at nada.

Higher derivatives [edit]

Faà di Bruno's formula generalizes the chain rule to higher derivatives. Assuming that y = f(u) and u = chiliad(x), then the first few derivatives are:

${\displaystyle {\begin{aligned}{\frac {dy}{dx}}&={\frac {dy}{du}}{\frac {du}{dx}}\\[4pt]{\frac {d^{2}y}{dx^{2}}}&={\frac {d^{ii}y}{du^{ii}}}\left({\frac {du}{dx}}\right)^{2}+{\frac {dy}{du}}{\frac {d^{2}u}{dx^{two}}}\\[4pt]{\frac {d^{3}y}{dx^{three}}}&={\frac {d^{3}y}{du^{3}}}\left({\frac {du}{dx}}\correct)^{three}+3\,{\frac {d^{2}y}{du^{2}}}{\frac {du}{dx}}{\frac {d^{2}u}{dx^{2}}}+{\frac {dy}{du}}{\frac {d^{3}u}{dx^{iii}}}\\[4pt]{\frac {d^{four}y}{dx^{4}}}&={\frac {d^{four}y}{du^{4}}}\left({\frac {du}{dx}}\right)^{4}+6\,{\frac {d^{3}y}{du^{iii}}}\left({\frac {du}{dx}}\right)^{ii}{\frac {d^{2}u}{dx^{2}}}+{\frac {d^{2}y}{du^{ii}}}\left(four\,{\frac {du}{dx}}{\frac {d^{3}u}{dx^{3}}}+three\,\left({\frac {d^{ii}u}{dx^{2}}}\right)^{2}\right)+{\frac {dy}{du}}{\frac {d^{four}u}{dx^{4}}}.\end{aligned}}}$

Proofs [edit]

Beginning proof [edit]

I proof of the chain rule begins by defining the derivative of the composite function f ∘ g , where we take the limit of the deviation quotient for f ∘ g as x approaches a:

${\displaystyle (f\circ m)'(a)=\lim _{ten\to a}{\frac {f(g(x))-f(g(a))}{x-a}}.}$

Assume for the moment that ${\displaystyle grand(x)\!}$ does not equal ${\displaystyle g(a)}$ for whatever x nigh a. Then the previous expression is equal to the product of two factors:

${\displaystyle \lim _{ten\to a}{\frac {f(g(x))-f(thou(a))}{g(x)-g(a)}}\cdot {\frac {g(x)-chiliad(a)}{x-a}}.}$

If ${\displaystyle m}$ oscillates about a, then it might happen that no thing how shut one gets to a, there is always an even closer x such that one thousand(ten) = g(a). For example, this happens virtually a = 0 for the continuous part g divers past g(ten) = 0 for 10 = 0 and g(x) = x ² sin(1/x) otherwise. Whenever this happens, the above expression is undefined because it involves segmentation by zilch. To work around this, introduce a function ${\displaystyle Q}$ as follows:

${\displaystyle Q(y)={\brainstorm{cases}\displaystyle {\frac {f(y)-f(g(a))}{y-g(a)}},&y\neq g(a),\\f'(g(a)),&y=one thousand(a).\end{cases}}}$

We volition show that the difference caliber for f ∘ one thousand is always equal to:

${\displaystyle Q(yard(10))\cdot {\frac {g(x)-grand(a)}{10-a}}.}$

Whenever g(x) is not equal to g(a), this is clear because the factors of thousand(x) − k(a) cancel. When g(10) equals g(a), then the difference quotient for f ∘ g is nil considering f(g(10)) equals f(g(a)), and the in a higher place product is null because it equals f′(g(a)) times zip. And then the to a higher place production is always equal to the difference quotient, and to show that the derivative of f ∘ g at a exists and to decide its value, we demand only testify that the limit as x goes to a of the to a higher place product exists and determine its value.

To exercise this, call back that the limit of a product exists if the limits of its factors exist. When this happens, the limit of the product of these two factors will equal the product of the limits of the factors. The two factors are Q(1000(10)) and (g(ten) − g(a)) / (ten − a). The latter is the difference caliber for g at a, and because g is differentiable at a by assumption, its limit every bit 10 tends to a exists and equals chiliad′(a).

As for Q(g(x)), notice that Q is defined wherever f is. Furthermore, f is differentiable at one thousand(a) by assumption, so Q is continuous at g(a), by definition of the derivative. The function g is continuous at a because information technology is differentiable at a, and therefore Q ∘ g is continuous at a. And so its limit every bit ten goes to a exists and equals Q(thousand(a)), which is f′(grand(a)).

This shows that the limits of both factors exist and that they equal f′(m(a)) and one thousand′(a), respectively. Therefore, the derivative of f ∘ g at a exists and equals f′(1000(a)) g′(a).

Second proof [edit]

Another way of proving the chain rule is to measure the fault in the linear approximation determined by the derivative. This proof has the advantage that it generalizes to several variables. It relies on the following equivalent definition of differentiability at a point: A function thou is differentiable at a if at that place exists a real number g′(a) and a office ε(h) that tends to zero equally h tends to nada, and furthermore

${\displaystyle g(a+h)-yard(a)=g'(a)h+\varepsilon (h)h.}$

Here the left-paw side represents the true deviation between the value of g at a and at a + h , whereas the correct-hand side represents the approximation determined by the derivative plus an error term.

In the situation of the chain rule, such a function ε exists because g is assumed to be differentiable at a. Again past supposition, a similar part besides exists for f at grand(a). Calling this function η, we have

${\displaystyle f(g(a)+k)-f(thou(a))=f'(g(a))k+\eta (one thousand)g.}$

The above definition imposes no constraints on η(0), even though it is assumed that η(k) tends to zero as k tends to zero. If we fix η(0) = 0, then η is continuous at 0.

Proving the theorem requires studying the difference f(chiliad(a + h)) − f(g(a)) as h tends to zip. The kickoff footstep is to substitute for g(a + h) using the definition of differentiability of 1000 at a:

${\displaystyle f(g(a+h))-f(g(a))=f(g(a)+g'(a)h+\varepsilon (h)h)-f(g(a)).}$

The side by side footstep is to use the definition of differentiability of f at g(a). This requires a term of the course f(g(a) + k) for some k. In the higher up equation, the correct m varies with h. Set g _h = thou′(a) h + ε(h) h and the right hand side becomes f(g(a) + g _h ) − f(g(a)). Applying the definition of the derivative gives:

${\displaystyle f(g(a)+k_{h})-f(k(a))=f'(g(a))k_{h}+\eta (k_{h})k_{h}.}$

To written report the behavior of this expression as h tends to aught, expand thou _h . After regrouping the terms, the right-hand side becomes:

${\displaystyle f'(thou(a))k'(a)h+[f'(g(a))\varepsilon (h)+\eta (k_{h})g'(a)+\eta (k_{h})\varepsilon (h)]h.}$

Because ε(h) and η(yard _h ) tend to zero as h tends to zippo, the first two bracketed terms tend to zero equally h tends to zero. Applying the same theorem on products of limits as in the first proof, the 3rd bracketed term also tends zero. Because the to a higher place expression is equal to the deviation f(one thousand(a + h)) − f(g(a)), by the definition of the derivative f ∘ g is differentiable at a and its derivative is f′(1000(a)) yard′(a).

The role of Q in the first proof is played by η in this proof. They are related by the equation:

${\displaystyle Q(y)=f'(thou(a))+\eta (y-chiliad(a)).}$

The need to define Q at g(a) is coordinating to the demand to define η at zero.

Third proof [edit]

Constantin Carathéodory'south alternative definition of the differentiability of a office can exist used to give an elegant proof of the concatenation rule.^[four]

Under this definition, a part f is differentiable at a point a if and only if there is a office q, continuous at a and such that f(x) − f(a) = q(ten)(x − a). At that place is at most one such function, and if f is differentiable at a and so f ′(a) = q(a).

Given the assumptions of the chain dominion and the fact that differentiable functions and compositions of continuous functions are continuous, nosotros have that there exist functions q, continuous at g(a), and r, continuous at a, and such that,

${\displaystyle f(m(x))-f(g(a))=q(g(10))(g(x)-m(a))}$

and

${\displaystyle g(x)-grand(a)=r(10)(x-a).}$

Therefore,

${\displaystyle f(g(x))-f(k(a))=q(g(x))r(x)(x-a),}$

simply the function given by h(ten) = q(thousand(x))r(x) is continuous at a, and we become, for this a

${\displaystyle (f(g(a)))'=q(g(a))r(a)=f'(g(a))yard'(a).}$

A like approach works for continuously differentiable (vector-)functions of many variables. This method of factoring also allows a unified approach to stronger forms of differentiability, when the derivative is required to be Lipschitz continuous, Hölder continuous, etc. Differentiation itself tin can be viewed as the polynomial remainder theorem (the little Bézout theorem, or factor theorem), generalized to an appropriate grade of functions.^{[ commendation needed ]}

Proof via infinitesimals [edit]

If ${\displaystyle y=f(x)}$ and ${\displaystyle x=g(t)}$ then choosing infinitesimal ${\displaystyle \Delta t\not =0}$ we compute the corresponding ${\displaystyle \Delta x=g(t+\Delta t)-chiliad(t)}$ and and then the respective ${\displaystyle \Delta y=f(x+\Delta x)-f(x)}$ , so that

${\displaystyle {\frac {\Delta y}{\Delta t}}={\frac {\Delta y}{\Delta 10}}{\frac {\Delta x}{\Delta t}}}$

and applying the standard part we obtain

${\displaystyle {\frac {dy}{dt}}={\frac {dy}{dx}}{\frac {dx}{dt}}}$

which is the chain dominion.

Multivariable case [edit]

The generalization of the chain rule to multi-variable functions is rather technical. Withal, information technology is simpler to write in the case of functions of the class

${\displaystyle f(g_{one}(x),\dots ,g_{g}(ten)).}$

As this case occurs often in the written report of functions of a single variable, it is worth describing it separately.

Case of f(one thousand ₁(ten), ... , g _k (ten)) [edit]

For writing the concatenation rule for a role of the class

f(g ₁(10), ... , g _k (ten)),

i needs the partial derivatives of f with respect to its chiliad arguments. The usual notations for partial derivatives involve names for the arguments of the function. As these arguments are non named in the above formula, it is simpler and clearer to announce by

${\displaystyle D_{i}f}$

the fractional derivative of f with respect to its ithursday argument, and by

${\displaystyle D_{i}f(z)}$

the value of this derivative at z.

With this notation, the concatenation dominion is

${\displaystyle {\frac {d}{dx}}f(g_{1}(x),\dots ,g_{thousand}(ten))=\sum _{i=1}^{k}\left({\frac {d}{dx}}{g_{i}}(x)\correct)D_{i}f(g_{ane}(10),\dots ,g_{grand}(x)).}$

Case: arithmetics operations [edit]

If the part f is addition, that is, if

${\displaystyle f(u,five)=u+v,}$

and then ${\textstyle D_{1}f={\frac {\partial f}{\partial u}}=1}$ and ${\textstyle D_{2}f={\frac {\partial f}{\partial v}}=1}$ . Thus, the chain rule gives

${\displaystyle {\frac {d}{dx}}(g(x)+h(ten))=\left({\frac {d}{dx}}g(x)\right)D_{one}f+\left({\frac {d}{dx}}h(x)\right)D_{2}f={\frac {d}{dx}}yard(x)+{\frac {d}{dx}}h(x).}$

For multiplication

${\displaystyle f(u,v)=uv,}$

the partials are ${\displaystyle D_{1}f=five}$ and ${\displaystyle D_{2}f=u}$ . Thus,

${\displaystyle {\frac {d}{dx}}(g(x)h(x))=h(10){\frac {d}{dx}}g(x)+g(ten){\frac {d}{dx}}h(x).}$

The example of exponentiation

${\displaystyle f(u,five)=u^{v}}$

is slightly more than complicated, every bit

${\displaystyle D_{1}f=vu^{v-1},}$

and, as ${\displaystyle u^{v}=e^{v\ln u},}$

${\displaystyle D_{two}f=u^{5}\ln u.}$

It follows that

${\displaystyle {\frac {d}{dx}}\left(g(x)^{h(x)}\right)=h(x)g(x)^{h(x)-1}{\frac {d}{dx}}1000(x)+g(10)^{h(x)}\ln 1000(x){\frac {d}{dx}}h(x).}$

Full general dominion [edit]

The simplest fashion for writing the chain dominion in the full general case is to use the total derivative, which is a linear transformation that captures all directional derivatives in a single formula. Consider differentiable functions f : R ^k → R ^k and yard : R ⁿ → R ^{one thousand} , and a point a in R ⁿ . Allow D _a g denote the total derivative of g at a and D _g(a) f denote the total derivative of f at g(a). These 2 derivatives are linear transformations R ⁿ → R ^m and R ^m → R ^k , respectively, and then they can be composed. The chain rule for total derivatives is that their composite is the full derivative of f ∘ g at a :

${\displaystyle D_{\mathbf {a} }(f\circ g)=D_{chiliad(\mathbf {a} )}f\circ D_{\mathbf {a} }g,}$

or for curt,

${\displaystyle D(f\circ g)=Df\circ Dg.}$

The higher-dimensional chain rule can be proved using a technique like to the 2d proof given to a higher place.^[v]

Because the total derivative is a linear transformation, the functions appearing in the formula can exist rewritten as matrices. The matrix respective to a total derivative is called a Jacobian matrix, and the composite of two derivatives corresponds to the product of their Jacobian matrices. From this perspective the concatenation rule therefore says:

${\displaystyle J_{f\circ g}(\mathbf {a} )=J_{f}(k(\mathbf {a} ))J_{g}(\mathbf {a} ),}$

or for short,

${\displaystyle J_{f\circ g}=(J_{f}\circ k)J_{g}.}$

That is, the Jacobian of a composite function is the production of the Jacobians of the composed functions (evaluated at the advisable points).

The college-dimensional chain rule is a generalization of the one-dimensional chain rule. If thousand, m, and n are one, so that f : R → R and g : R → R , then the Jacobian matrices of f and m are 1 × 1. Specifically, they are:

${\displaystyle {\brainstorm{aligned}J_{thou}(a)&={\begin{pmatrix}m'(a)\end{pmatrix}},\\J_{f}(yard(a))&={\begin{pmatrix}f'(thou(a))\end{pmatrix}}.\end{aligned}}}$

The Jacobian of f ∘ grand is the product of these 1 × one matrices, so it is f′(thousand(a))⋅k′(a), as expected from the one-dimensional chain rule. In the language of linear transformations, D _a (g) is the function which scales a vector past a factor of g′(a) and D _g(a)(f) is the function which scales a vector by a factor of f′(g(a)). The chain rule says that the composite of these two linear transformations is the linear transformation D _a (f ∘ chiliad), and therefore information technology is the function that scales a vector by f′(1000(a))⋅m′(a).

Some other style of writing the chain dominion is used when f and chiliad are expressed in terms of their components as y = f(u) = (f _one(u), …, f _k (u)) and u = g(x) = (thou _one(ten), …, g _grand (x)). In this case, the in a higher place rule for Jacobian matrices is normally written as:

${\displaystyle {\frac {\fractional (y_{1},\ldots ,y_{m})}{\fractional (x_{ane},\ldots ,x_{n})}}={\frac {\fractional (y_{1},\ldots ,y_{g})}{\fractional (u_{1},\ldots ,u_{m})}}{\frac {\partial (u_{1},\ldots ,u_{m})}{\partial (x_{i},\ldots ,x_{n})}}.}$

The chain dominion for full derivatives implies a chain rule for partial derivatives. Recall that when the total derivative exists, the partial derivative in the ithursday coordinate management is found by multiplying the Jacobian matrix past the ithursday ground vector. By doing this to the formula higher up, we find:

${\displaystyle {\frac {\partial (y_{1},\ldots ,y_{grand})}{\partial x_{i}}}={\frac {\fractional (y_{1},\ldots ,y_{1000})}{\fractional (u_{one},\ldots ,u_{yard})}}{\frac {\partial (u_{1},\ldots ,u_{thousand})}{\partial x_{i}}}.}$

Since the entries of the Jacobian matrix are partial derivatives, nosotros may simplify the to a higher place formula to become:

${\displaystyle {\frac {\partial (y_{1},\ldots ,y_{g})}{\fractional x_{i}}}=\sum _{\ell =ane}^{m}{\frac {\partial (y_{1},\ldots ,y_{k})}{\fractional u_{\ell }}}{\frac {\partial u_{\ell }}{\partial x_{i}}}.}$

More than conceptually, this dominion expresses the fact that a alter in the x _i direction may change all of g _one through grand_m , and whatsoever of these changes may affect f.

In the special example where yard = i, so that f is a real-valued role, so this formula simplifies even further:

${\displaystyle {\frac {\partial y}{\fractional x_{i}}}=\sum _{\ell =one}^{1000}{\frac {\partial y}{\fractional u_{\ell }}}{\frac {\partial u_{\ell }}{\fractional x_{i}}}.}$

This can exist rewritten as a dot product. Recalling that u = (g _one, …, g _thousand ), the fractional derivative ∂u / ∂ten _i is also a vector, and the chain dominion says that:

${\displaystyle {\frac {\partial y}{\fractional x_{i}}}=\nabla y\cdot {\frac {\partial \mathbf {u} }{\partial x_{i}}}.}$

Instance [edit]

Given u(x, y) = 10 ² + 2y where 10(r, t) = r sin(t) and y(r,t) = sin²(t), determine the value of ∂u / ∂r and ∂u / ∂t using the concatenation rule.

${\displaystyle {\frac {\partial u}{\partial r}}={\frac {\partial u}{\partial x}}{\frac {\partial x}{\partial r}}+{\frac {\fractional u}{\partial y}}{\frac {\partial y}{\fractional r}}=(2x)(\sin(t))+(2)(0)=2r\sin ^{2}(t),}$

and

${\displaystyle {\begin{aligned}{\frac {\partial u}{\partial t}}&={\frac {\fractional u}{\partial x}}{\frac {\partial x}{\fractional t}}+{\frac {\fractional u}{\partial y}}{\frac {\partial y}{\fractional t}}\\&=(2x)(r\cos(t))+(two)(2\sin(t)\cos(t))\\&=(2r\sin(t))(r\cos(t))+4\sin(t)\cos(t)\\&=2(r^{2}+2)\sin(t)\cos(t)\\&=(r^{2}+2)\sin(2t).\cease{aligned}}}$

Higher derivatives of multivariable functions [edit]

Faà di Bruno'southward formula for college-order derivatives of single-variable functions generalizes to the multivariable case. If y = f(u) is a function of u = g(x) as above, so the second derivative of f ∘ thou is:

${\displaystyle {\frac {\partial ^{ii}y}{\partial x_{i}\fractional x_{j}}}=\sum _{grand}\left({\frac {\partial y}{\partial u_{k}}}{\frac {\fractional ^{2}u_{k}}{\partial x_{i}\fractional x_{j}}}\correct)+\sum _{k,\ell }\left({\frac {\partial ^{2}y}{\fractional u_{thou}\partial u_{\ell }}}{\frac {\partial u_{1000}}{\fractional x_{i}}}{\frac {\partial u_{\ell }}{\partial x_{j}}}\right).}$

Further generalizations [edit]

All extensions of calculus have a concatenation rule. In most of these, the formula remains the aforementioned, though the meaning of that formula may exist vastly different.

One generalization is to manifolds. In this situation, the chain rule represents the fact that the derivative of f ∘ g is the composite of the derivative of f and the derivative of g. This theorem is an immediate event of the higher dimensional chain rule given in a higher place, and it has exactly the aforementioned formula.

The chain dominion is likewise valid for Fréchet derivatives in Banach spaces. The same formula holds as before.^[half-dozen] This case and the previous one admit a simultaneous generalization to Banach manifolds.

In differential algebra, the derivative is interpreted as a morphism of modules of Kähler differentials. A band homomorphism of commutative rings f : R → S determines a morphism of Kähler differentials Df : Ω _R → Ω _S which sends an element dr to d(f(r)), the exterior differential of f(r). The formula D(f ∘ thou) = Df ∘ Dg holds in this context as well.

The common characteristic of these examples is that they are expressions of the idea that the derivative is part of a functor. A functor is an operation on spaces and functions between them. It associates to each space a new space and to each function betwixt two spaces a new function between the corresponding new spaces. In each of the above cases, the functor sends each space to its tangent bundle and it sends each role to its derivative. For instance, in the manifold case, the derivative sends a C ^r -manifold to a C ^r−1-manifold (its tangent parcel) and a C ^r -office to its total derivative. In that location is 1 requirement for this to be a functor, namely that the derivative of a composite must exist the composite of the derivatives. This is exactly the formula D(f ∘ g) = Df ∘ Dg .

There are besides concatenation rules in stochastic calculus. Ane of these, Itō's lemma, expresses the blended of an Itō process (or more generally a semimartingale) dX _t with a twice-differentiable function f. In Itō'southward lemma, the derivative of the composite office depends not only on dX _t and the derivative of f but also on the second derivative of f. The dependence on the second derivative is a result of the not-zero quadratic variation of the stochastic process, which broadly speaking means that the process can move upward and down in a very rough mode. This variant of the chain rule is not an instance of a functor because the two functions beingness composed are of dissimilar types.

See also [edit]

• Integration by exchange
• Leibniz integral rule
• Quotient rule
• Triple production rule
• Product rule
• Automatic differentiation, a computational method that makes heavy use of the chain rule to compute exact numerical derivatives.

References [edit]

1. ^ George F. Simmons, Calculus with Analytic Geometry (1985), p. 93.
2. ^ Rodríguez, Omar Hernández; López Fernández, Jorge Yard. (2010). "A Semiotic Reflection on the Pedagogy of the Chain Rule". The Mathematics Enthusiast. vii (2): 321–332. Retrieved 2019-08-04 .
3. ^ Apostol, Tom (1974). Mathematical assay (2d ed.). Addison Wesley. Theorem five.v.
4. ^ Kuhn, Stephen (1991). "The Derivative á la Carathéodory". The American Mathematical Monthly. 98 (1): 40–44. JSTOR 2324035.
5. ^ Spivak, Michael (1965). Calculus on Manifolds. Boston: Addison-Wesley. pp. 19–twenty. ISBN0-8053-9021-9.
6. ^ Cheney, Ward (2001). "The Chain Dominion and Mean Value Theorems". Assay for Applied Mathematics. New York: Springer. pp. 121–125. ISBN0-387-95279-ix.

External links [edit]

• "Leibniz rule", Encyclopedia of Mathematics, Ems Press, 2001 [1994]
• Weisstein, Eric Due west. "Chain Rule". MathWorld.

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Source: https://en.wikipedia.org/wiki/Chain_rule

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