Differentiation under the integral sign, please help

Question

Find the partial derivatives of the function: $$\int_{x^2e^{5y}}^{\ln(x^3-2)}\cos(t^2)dt$$

Maple responds: $$-2\,\cos \left( {x}^{4} \left( {{\rm e}^{5\,y}} \right) ^{2} \right) x {{\rm e}^{5\,y}}+3\,{\frac {\cos \left( \left( \ln \left( {x}^{3}-2 \right) \right) ^{2} \right) {x}^{2}}{{x}^{3}-2}}.$$ What's the steps to solve this? I'm having trouble applying the FTC with two variables and partial derivatives...

Answer 1

The steps to differentiate under the integral sign are as follows; \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} &\left (\int_{a(x)}^{b(x)}f(x,t)\,\mathrm{d}t \right)= \\ &\quad= f\big(x,b(x)\big)\cdot b'(x) - f\big(x,a(x)\big)\cdot a'(x) + \int_{a(x)}^{b(x)} f_x(x,t)\; \mathrm{d}t. \end{align} Basically, the limits are important as you must evaluate your function at these limits and evaluate a partial derviative in the last term, that's what the subsrcipt $x$ means.

For reference, see the Leibniz Integral Rule on Wolfram Mathworld.

Answer 2

You can write $$f(x,y) = \int_0^{\ln(x^3 - 2)} \cos(t^2) \, dt - \int_0^{x^2 e^{5y}} \cos(t^2) \, dt.$$ Then use the fact that $$\frac{d}{dz} \int_0^{h(z)} \cos(t^2) \, dt = \cos(h(z))^2 h'(z)$$according to the fundamental theorem.