I was requested to find the derivative of $f(x)$ and $g(x)$, which are defined as follows:
$$f(x)= \int_0 ^{x^2} \frac{e^{t^2} +1}{\sqrt{1-t^2}} dt$$
$$ g(x)=\int_\sqrt{x} ^{x^3} \frac{t+1}{\sqrt{1+2^t}} dt$$
I was able to find derivatives of integral functions whose integration bounds were $a$ to $x$, using the fundamental theorem of calculus, but this exponentiated bounds are killing me. How should one proceed to solve such a problem?
Best Answer
By the fundamental theorem of calculus, $$\frac{d}{dx}\int_{u(x)}^{v(x)}h(t) dt=h\big(v(x)\big)v'(x)-h\big(u(x)\big)u'(x).$$
So $$\frac{d}{dx}\int_{0}^{x^2}h(t) dt=h\big(x^2\big)\cdot 2x-h\big(0\big)\cdot0.=2xh(x^2)$$
$$\frac{d}{dx}\int_{\sqrt x}^{x^3}h(t) dt=h\big(x^3\big)\cdot 3x^2-h\big(\sqrt x\big)\cdot \frac{1}{2\sqrt x}=3x^2h(x^3)-\frac{1}{2\sqrt x}h\big(\sqrt x\big).$$