- #1

Krushnaraj Pandya

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## Homework Statement

if ## f(x) ={\int_{\frac{\pi^2}{16}}^{x^2}} \frac {\cos x \cos \sqrt{z}}{1+\sin^2 \sqrt{z}} dz## then find ## f'(\pi)##

**2. The given solution**

Differentiating both sides w.r.t x

##f'(x) = {-\sin x {\int_{\frac{\pi^2}{16}}^{x^2}} \frac{\cos \sqrt{z}}{1+\sin^2 \sqrt{z}} dz }+{ \frac{\cos x \cos x}{1+\sin^2{x}} 2x } - {0} ##

then put ##\pi## in place of x to find the answer ##2 \pi##

**3. The problems in the solution**

Note- I have the solution but am unable to understand it. I encountered this while trying to learn the application of newton-leibniz theorem and I'm comfortable with its basic application when integral is in the form ## \int_{n(x)}^{g(x)} f(x) dx##. I'm having trouble understanding

1) there are two variables, z and x wherein I have seen the theorem being applied only for a single variable everywhere.

2) The term after the plus sign is clearly the same procedure as in the theorem where the upper limit ##x^2## was put in place of z but it was not put in place of cosx as cos(x^2) but only for the rest of the function where z is present.

3) No idea how we got the first term. all I know is they differentiated cosx and somehow took -sinx out of the integral sign

maybe they used product rule of derivatives but is that possible under the integral sign?

Please help me learn this better and remove my conceptual doubts- I'd be really grateful. Thank you.