Some Fundamental theory of calc questions I just wanted to sanity check these questions:


  Find the derivative of these functions



$$g(s) = \int_{5}^s (t - t^2) ^8 dt$$


$$ g'(s) = (t - t^2)^8$$


$$h(x) = \int_{1}^\sqrt{x} \frac{z^2}{z^4 +1} dz$$


$$h'(x) = \frac{1}{2\sqrt{x}} \cdot \frac{z^2}{z^4 + 1}$$
For the first one just replace $t$ with $s$ in the integrand. 

For the second I suppose the chain rule gives:  $$h'(x)=\frac1{2\sqrt x}\cdot \frac x{x^2+1}$$.
The first should be $g'(s)=(s-s^2)^8$  and the second is  $h'(x)=\frac1{2\sqrt x}\cdot \frac x{x^2+1}$

This follows from the Leibniz integral rule of differentiation under the integral sign  , which is,

$\,\frac{d}{dx}\bigg(\huge \int_{\small a(x)}^{\small b(x)}\large f(x,t)\,dt\bigg) =\small f(x,b(x)).\frac{d(b(x))}{dx}-f(x,a(x)).\frac{d(a(x))}{dx}+\int_{\small{a(x)}}^{\small{b(x)}}\partial_xf(x,t)dt$.

In your case since the partial derivatives $\,0$  you can just omit them. 

EDIT:

you are doing $\frac{d}{dx}(\int_1^{\sqrt x}\frac{z^2}{z^4+1})$. Thus  you sub in $z=\sqrt x$ after you differentiate(because of the above mentioned  rule) You get $\frac{(\sqrt x)^2}{(\sqrt x)^4+1}\cdot\frac1{2\sqrt x} = \frac{x}{x^2+1}\cdot\frac1{2\sqrt x}$. Also you dont get the term $\frac{1^2}{1^4+1}$ as $\frac{d}{dx}(1) =0$. Cancelling  that term.