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I was using the chain rule and i have $F'(x) = H(x, g(x)) \cdot g'(x)$. Is this right?

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this may help – newbie Aug 18 '11 at 15:29
Put $$A(u,v):=\int_a^{g(u)}H(v,t)\ dt$$ and note that we have $$F’(x)=\frac{\partial A}{\partial u}(x,x)+\frac{\partial A}{\partial v}(x,x)$$ by the Chain Rule. Do you agree? – Pierre-Yves Gaillard Aug 18 '11 at 15:34
See the question How do I differentiate this integral? too. – Américo Tavares Aug 22 '11 at 11:52
up vote 2 down vote accepted

For a correct formula, see Plop's answer. However, you should always try to verify your formula using concrete examples, and see if your formula works.

For instance, let's pick a $H(x,t)$ and a $g(x)$ and see what happens. Let's pick $H(x,t)=x+t$, $g(x)= x $ and $a=0$. Then:

$$F(x) = \int_a^{x} (x+t)dt = \left( xt + \frac{t^2}{2}\right)_0^{x} = x(x)+\frac{(x)^2}{2} = \frac{3x^2}{2}.$$ Thus, $F'(x) = 3x$. According to your formula, this should have been $$H(x,g(x))\cdot g'(x)=(x+x)\cdot 1=2x \neq 3x = F'(x).$$ In the notation of the correct formula given by Plop, we have $G(x,y)=\int_0^y (x+t)dt$ and $F(x)=G(x,x)$, which, in this case, it is $$G(x,y)=\int_0^y (x+t)dt = \left( xt + \frac{t^2}{2}\right)_0^{y}=xy+\frac{y^2}{2},$$ $$\frac{\partial G}{\partial x}(x,y) = y,\quad \frac{\partial G}{\partial y}(x,y) = x+y.$$ Thus, the formula reads: $$F'(x) = \frac{\partial G}{\partial x}(x,g(x))+g'(x)\frac{\partial G}{\partial y}(x,g(x)) = x + 1\cdot (x+x) = x+2x=3x,$$ which is indeed correct.

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No. Let $G(x,y)=\int_a^y H(x,t) dt$. Then $F(x)=G(x,g(x))$ and thus (by the chain rule in 2 variables) $F'(x)=\frac{\partial G}{\partial x}(x,g(x))+g'(x)\frac{\partial G}{\partial y}(x,g(x))$ so you need to compute these two partial derivatives. The one wrt $y$ is given by the so-called "fundamental theorem of analysis", while the other is differentiation under the integral sign (assuming $H$ is regular enough).

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Since there have been several questions of this kind, it might be worth recalling the Chain Rule. I'll use a definition of differentiability which is due to Carathéodory, and which is of course equivalent to the usual one.

Carathéodory's definition of differentiability

Definition 1. Let $a$ be a point of an open subset $U$ of $\mathbb R^n$ and $f:U\to\mathbb R^k$ a function. Then $f$ is differentiable at $a$ if there is a function $\varphi$ from $U$ to the space of linear maps from $\mathbb R^n$ to $\mathbb R^k$, which is continuous at $a$, and satisfies $$f(x)=f(a)+\varphi(x)(x-a)$$ for all $x$ in $U$.

We claim that, in this situation, the linear map $\varphi(a)$ doesn't depend on $\varphi$. To see this, let $\psi$ be a function satisfying the same assumptions as $\varphi$, let $v$ be any vector of $\mathbb R^n$, and let $\varepsilon > 0$ be small enough to ensure that $x:=a+tv$ is in $U$ whenever $|t| < \varepsilon$. An easy computation shows $\varphi(a+tv)v=\psi(a+tv)v$ for $0 < |t| < \varepsilon$. As $\varphi$ and $\psi$ are continuous at $a$, this implies $\varphi(a)v=\psi(a)v$, and thus, $v$ being arbitrary, $\varphi(a)=\psi(a)$. This justifies the following definition.

Definition 2. In the setting of Definition 1, we put $f'(a):=\varphi(a)$, and call this linear map the derivative of $f$ at $a$.

The Chain Rule

Let now $b$ be a point of an open subset $V$ of $\mathbb R^k$ and $g:V\to\mathbb R^m$ a function. Assume that $f(U)$ is contained in $V$, that $f(a)$ is equal to $b$, that $f$ is differentiable at $a$, and that $g$ is differentiable at $b$. Then $g\circ f$ is differentiable at $a$, and we have $$(g\circ f)'(a)=g'(f(a))\circ f'(a).$$

Proof. By assumption we have $g(y)=g(b)+\psi(y)(y-b)$. This gives $$g(f(x))=g(b)+\psi(f(x))(f(x)-b)$$ $$=g(b)+\psi(f(x))\Big(\varphi(x)(x-a)\Big)$$ $$=g(b)+\Big(\psi(f(x))\circ\varphi(x)\Big)(x-a).$$

Particular cases

Let's look at three more and more particular cases.

(1) If $n=1$ then
$$\frac{d}{dt}\ g\Big(f_1(t),\dots,f_k(t)\Big)=\sum_{i=1}^k\ \frac{\partial g}{\partial x_i}\Big(f_1(t),\dots,f_k(t)\Big)f_i'(t).$$

(2) If in addition $f(t)=ta$, $a=(a_1,\dots,a_k)\in\mathbb R^k$, then $$\frac{d}{dt}\ g(ta)=\sum_{i=1}^k\ a_i\ \frac{\partial g}{\partial x_i}(ta).$$

(3) If in addition $a_i=1$ for all $i$, then $$\frac{d}{dt}\ g(t,\dots,t)=\sum_{i=1}^k\ \frac{\partial g}{\partial x_i}(t,\dots,t).$$


(JSTOR) The Derivative à la Carathéodory, Stephen Kuhn, The American Mathematical Monthly, Vol. 98, No. 1 (Jan., 1991), pp. 40-44.

[For the question stricto sensu, see the other answers.]


It is of course important to check that Carathéodory's definition of differentiability is equivalent to the usual one. Let us do it briefly. It is clear that Carathéodory's definition implies the usual one. Let us verify the converse. We can assume $0\in U\subset\mathbb R^n$, $f:U\to\mathbb R$, and $f(0)=0=f'(0)$. Our assumption is $f(x)=|x|\varepsilon(x)$ where $\varepsilon(x)$ tends to $0$ as $x$ tends to $0$. We must write $f(x)$ in the form $\varphi(x)\cdot x$, where $u\cdot v$ is the standard inner product on $\mathbb R^n$, and where $\varphi(x)$ is in $\mathbb R^n$ and tends to $0$ as $x$ tends to $0$. It suffices to define $\varphi(x)$ for $x\not=0$. One easily sees that the formula $\varphi(x):=\varepsilon(x)|x|^{-1}x$ works.

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Nice article there! I am out of votes at the moment, but I shall upvote when I can do so again... – J. M. Aug 21 '11 at 16:47
This is probably the favourite answer I've read on Stackexchange yet :-) So unbelievably natural, thank you! – Eivind Dahl Oct 14 '11 at 17:22

Hint Think of $F(x) = G(g(x),x)$ where $G(u,x) = \int_a^u H(x,t)\ dt$, and use a two-variable version of the chain rule.

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Thanks everyone. I'm novice in multi-variable calculus, but with your hints I'm glad. – Mario De León Urbina Aug 18 '11 at 16:15

This is known as the Leibniz rule.

You can also do it by definition:

$$F(x+\delta) = \int_{a}^{g(x+\delta)} H(x+\delta,t) dt$$

$$F(x+\delta) - F(x) = \int_{a}^{g(x)} [ H(x+\delta,t) - H(x,t)] dt + \int_{g(x)}^{g(x+\delta)} H(x+\delta,t) dt$$

$$ F'(x)=\lim_{\delta \to 0} \frac{F(x+\delta) - F(x)}{\delta} = \int_{a}^{g(x)} \frac{\partial H(x,t)}{\partial x} dt + H(x,g(x)) \; g'(x)$$

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Vote up coz I've mentioned Leibniz rule in the prior comment as well. :D – newbie Aug 18 '11 at 19:13

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