Some Fundamental theory of calc questions

Question

I just wanted to sanity check these questions:

Find the derivative of these functions

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

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

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

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

score 2 · Answer 1 · answered Mar 23 '18 at 18:54

2

For the first one just replace $t$ with $s$ in the integrand.

For the second I suppose the chain rule gives: $$h'(x)=\frac1{2\sqrt x}\cdot \frac x{x^2+1}$$.

answered Mar 23 '18 at 18:54

I guess I'm a little confused as to why I need to replace the t with the s. What is going on? – Jwan622 Mar 23 '18 at 18:56
That's the fundamental theorem of calculus... – Mar 23 '18 at 18:57
Don't just answer "it is what it is", explain why! – Dylan Mar 23 '18 at 19:26
@Jwan622 $t$ is just the dummy variable. They are replaced with $s$ in the resulting integral. This is why $g$ and $g'$ are functions of $s$ – Dylan Mar 23 '18 at 19:27
why does the right side reduce to $x/x^2+1$? – Jwan622 Apr 10 '18 at 17:39
By FTC, replace $z$ with $\sqrt x$. But don't forget the chain rule... – Apr 10 '18 at 19:27
$(\sqrt x)^2=x$ and $(\sqrt x)^4=x^2$... – Apr 10 '18 at 19:36

score 0 · Answer 2 · answered Mar 23 '18 at 18:33

0

Your first function $g'(s)$ still has $t$'s left over. I assume that once you replace them you will be OK.

Your second function $h'(x)$ still has $z$'s left over. I assume that once you replace them you will be OK.

answered Mar 23 '18 at 18:33

Valborg

181

user577215664 · Answer 3 · 2018-03-23T18:53:50.437

0

$g'(s)$ means that g is a function of the variable s but you still have t which is only valid inside the integrand... $$g'(s)=(s-s^2)^8$$

edited Mar 23 '18 at 18:53

answered Mar 23 '18 at 18:42

user577215664

40,625

The Integrator · Answer 4 · 2018-04-11T19:31:36.463

-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.

edited Apr 11 '18 at 19:31

answered Mar 24 '18 at 04:55

The Integrator

3,200

Why does the right side of the second one reduce to $x/x^2+1$? – Jwan622 Apr 10 '18 at 17:39
1

you are doing $\frac{d}{dx}(\int_1^{\sqrt x}\frac{z^2}{ z^4+1})$. Thus if you sub in $z= \sqrt x$ after you differentiate(because of the rule) You get $\frac{(\sqrt x)^2}{(\sqrt x)^4+1} = \frac{x}{x^2+1}$ – The Integrator Apr 10 '18 at 20:37
could you write this as the answer @pranavB23 and I'll give you credit? – Jwan622 Apr 11 '18 at 16:20

Some Fundamental theory of calc questions

4 Answers4