How do you derive definite integrals? How do you get $g'(y)$?
$$g(y)=\int_{1}^{e^{y^{2}}}\cos\left(\sqrt{\log(x)}\right)\,dx$$
 A: Well, in general:
$$\frac{\partial}{\partial\text{y}}\left\{\int_\text{a}^{\text{z}\left(\text{y}\right)}\text{f}\left(x\right)\space\text{d}x\right\}=\text{f}\left(\text{z}\left(\text{y}\right)\right)\cdot\text{z}\space'\left(\text{y}\right)\tag1$$
So, for your problem:
$$\frac{\partial}{\partial\text{y}}\left\{\int_1^{\exp\left(\text{y}^2\right)}\cos\left(\sqrt{\ln\left(x\right)}\right)\space\text{d}x\right\}=\cos\left(\sqrt{\ln\left(\exp\left(\text{y}^2\right)\right)}\right)\cdot\frac{\text{d}}{\text{d}\text{y}}\left(\exp\left(\text{y}^2\right)\right)=$$
$$2\cdot\text{y}\cdot\exp\left(\text{y}^2\right)\cdot\cos\left(\sqrt{\ln\left(\exp\left(\text{y}^2\right)\right)}\right)\tag2$$
And when $\text{y}$ is positive:
$$\sqrt{\ln\left(\exp\left(\text{y}^2\right)\right)}=\text{y}$$
A: We use the chain rule and the fundamental theorem of calculus to obtain
$$\begin{align}
g'(y)&=\frac{d}{dy}\int_0^{e^{y^2}} \cos\left(\sqrt{\log(x)}\right),dx\\\\
&=\frac{d}{d(e^{y^2})}\left(\int_0^{e^{y^2}} \cos\left(\sqrt{\log(x)}\right),dx\right)\,\left(\frac{de^{y^2}}{dy}\right)\\\\
&=\cos\left(\sqrt{\log(e^{y^2})}\right)\,\left(\frac{de^{y^2}}{dy}\right)\\\\
&=2ye^{y^2}\cos\left(y\right)
\end{align}$$
A: Let $f(x)=\cos(\sqrt{\ln x})$, and let $F(x)$ be its indefinite integral. Then
$$g(y)=F(e^{y^2})-F(1)$$
By taking the derivative of both sides and using the chain rule, we get
$$g'(y)=\cos(\sqrt{\ln (e^{y^2})})*\frac{d}{dy}(e^{y^2})$$
$$g'(y)=\cos(y)*2ye^{y^2}$$
Which should be the answer.
