# Does this equality always hold?

Is it true in general that $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x} f(u,x) \mathrm{d}u = \int_0^{x} \left( \frac{\mathrm{d}}{\mathrm{d}x} f(u,x) \right)\mathrm{d}u +f(x,x )$ ?

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Yes, it is always true see en.wikipedia.org/wiki/Differentiation_under_the_integral_sign. –  Fabian Jun 4 '11 at 10:02
@Fabian: thank you –  ghshtalt Jun 4 '11 at 11:08

Yes, it is, under the conditions indicated below.

Let $$I(x)=\displaystyle\int_{0}^{x}f(u,x)\; \mathrm{d}u.\qquad(\ast)$$

If $f(u,x)$ is a continuous function and $\partial f/\partial x$ exists and is continuous, then

$$I^{\prime }(x)=\displaystyle\int_{0}^{x}\dfrac{\partial f(u,x)}{\partial x}\; \mathrm{d}u+f(x,x)\qquad(\ast\ast)$$

follows from the Leibniz rule and chain rule.

Note: the integrand of $(\ast\ast)$ is a partial derivative.

It generalizes to the integral

$$I(x)=\displaystyle\int_{u(x)}^{v(x)}f(t,x)\; \mathrm{d}t.$$

Under suitable conditions ($u(x),v(x)$ are differentiable functions, $f(t,x)$ is a continuous function and $\partial f/\partial x$ exists and is continuous), we have

$$I^{\prime }(x)=\displaystyle\int_{u(x)}^{v(x)}\dfrac{\partial f(t,x)}{\partial x}\; \mathrm{d}t+f(v(x),x)v^{\prime }(x)-f(u(x),x)u^{\prime }(x).$$

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excellent, thank you –  ghshtalt Jun 4 '11 at 11:09
You are welcome! –  Américo Tavares Jun 4 '11 at 11:40
In short, it is true if $f$ and the partial derivative with respect to $f$ are continuous in the region of differentiation.