Leibniz's rule (derivatives and integrals)

See Leibniz's rule, a disambiguation page, for other meanings of this term.

---

In mathematics, Leibniz's rule for differentiation under the integral sign, named after Gottfried Leibniz, tells us that if we have an integral of the form

$\int _{y_{0}}^{y_{1}}f(x,y)\,dy$

then for $x\in (x_{0},x_{1})$ the derivative of this integral is thus expressible

${d \over dx}\,\int _{y_{0}}^{y_{1}}f(x,y)\,dy=\int _{y_{0}}^{y_{1}}{\partial \over \partial x}f(x,y)\,dy$

provided that $f$ and $\partial f/\partial x$ are both continuous over a region in the form

$[x_{0},x_{1}]\times [y_{0},y_{1}].$

Proof

The proof is straightforward: let us first make the assignment

$u(x)=\int _{y_{0}}^{y_{1}}f(x,y)\,dy.$

Then

${d \over dx}u(x)=\lim _{h\rightarrow 0}{u(x+h)-u(x) \over h}.$

Substituting back

$=\lim _{h\rightarrow 0}{\int _{y_{0}}^{y_{1}}f(x+h,y)\,dy-\int _{y_{0}}^{y_{1}}f(x,y)\,dy \over h}.$

Since integration is linear, we can write the two integrals as one:

$=\lim _{h\rightarrow 0}{\int _{y_{0}}^{y_{1}}(f(x+h,y)-f(x,y))\,dy \over h}.$

And we can take the constant inside, with the integrand

$=\lim _{h\rightarrow 0}\int _{y_{0}}^{y_{1}}{f(x+h,y)-f(x,y) \over h}\,dy.$

And now, since the integrand is in the form of a difference quotient:

$=\int _{y_{0}}^{y_{1}}{\partial \over \partial x}f(x,y)\,dy$

which can be justified by uniform continuity, so

$={d \over dx}u(x)={d \over dx}\int _{y_{0}}^{y_{1}}f(x,y)\,dy.$

Alternate form

For a monovariant function $g$:

${d \over dx}\,\int _{f_{1}(x)}^{f_{2}(x)}g(t)\,dt=g(f_{2}(x)){f_{2}'(x)}-g(f_{1}(x))f_{1}'(x)$