Question

Find the derivative.
d/dx
x^2
(integral) ln(7 + t^2) dt
0

Answer 1

FTC along with chain rule

Answer 2

\(\Large \dfrac{d}{dx}\int \limits_{f(x)}^{g(x)}h(x)dx = h(g(x))g'(x)-h(f(x))f'(x)
\)

Answer 3

\(\Large \dfrac{d}{dx}\int \limits_0^{g(x)}h(t)dt = h(g(t))g'(t)
\)

h(t) = ln(7+t^2)

Answer 4

g(x) = x^2

Answer 5

F(b)-F(a)… isn't that along the lines of what i could do? so i find the antiderivative and plug in those values right..?

Answer 6

\[
\frac{d}{dx}\int\limits_{0}^{x^2}\ln(7+t^2)~dt = \frac{d}{dx}\int\limits_{0}^{x^2}g'(t)~dt = \frac{d}{dx}\Bigg[g(x^2)-g(0)\Bigg]
\]

Answer 7

To solve:\[
\frac{d}{dx}\Bigg[g(x^2)-g(0)\Bigg]
\]You need to use the chain rule.

Answer 8

so it'd be 1/(7+t^2) * 2t

Answer 9

by the chain rule.. and then i plug in the values..?

Answer 10

Nope.

Answer 11

I didn't say \(g(t) = \ln(7+t^2)\). I said that \(g'(t) = \ln(7+t^2)\).

Answer 12

\(\Large \dfrac{d}{dx}\int \limits_0^{g(x)}h(t)dt = h(g(x))g'(x)\)
h(t) = ln (7+t^2)
g(x) = x^2

Answer 13

Also, the \(t\) went away when we integrated, so we won't be seeing it anymore.

Answer 14

okay… so i need to find the antiderivative of g'(t)… I'm sorry my professor ddnt demonstrate how the chain rule was used for these problems

Answer 15

i am so confused; im sorry

Answer 16

the anti derivative of g'(t) is just g(t)
which is shown by wio at this step
\(\frac{d}{dx}\int\limits_{0}^{x^2}\ln(7+t^2)~dt = \frac{d}{dx}\int\limits_{0}^{x^2}g'(t)~dt = \frac{d}{dx}\Bigg[g(x^2)-g(0)\Bigg]\)
in the end, he also plugged in the limits

Answer 17

\(\frac{d}{dx}\int\limits_{0}^{x^2}\ln(7+t^2)~dt = \frac{d}{dx}\int\limits_{0}^{x^2}g'(t)~dt =\frac{d}{dx} [g(t)]_{0}^{x^2}= \frac{d}{dx}\Bigg[g(x^2)-g(0)\Bigg]\)

Answer 18

and since g(0) is a constant, you just need
\(\Large \dfrac{d}{dx}g(x^2)\)
which will require chain rule

Answer 19

The 'trick' is that \[
\frac{d}{dx}g(x^2) = \frac{d}{d(x^2)}g(x^2)\frac{d}{dx}x^2 = g'(x^2)(2x)
\]

Answer 20

and we call that trick as chain rule

Answer 21

It is easier to see when you let \(u=x^2\), because then it is just: \[
\frac{dg}{dx}=\frac{dg}{du}\frac{du}{dx}
\]

Answer 22

I'm confused on why it requires the chain rule.. g(t) is composed with x^2 because of the whole F(b) -F(a), but i don't know why you just wouldn't stop at g(x^2). I understand how the chain rule works, I'm just not quite sure on why it takes place if that makes sense..?:/

Answer 23

So these problems would be easier if i use the technique of integration by substitution

Answer 24

The reason why we don't stop at \(g(t)\) is because we don't always know what \(g(t)\) is, but we know what \(g'(t)\) is, since it is the integrand.

Answer 25

oooohhh okay! so what would be the following step?

Answer 26

This was the furthest we got so far.\[
\frac{d}{dx}g(x^2) = \frac{d}{d(x^2)}g(x^2)\frac{d}{dx}x^2 = g'(x^2)\cdot 2x
\]

Answer 27

mhmm! i just don't know what to do with that. do we then apply the upper and lower limits?

Answer 28

Well, we want to get rid of the \(g'\), by remember what it originally was.

Answer 29

so g' was originally ln(7+t^2)

Answer 30

Yep, and we need to sub in \(x^2\) for \(t\).

Answer 31

so ln(7+(x^2)^2)

Answer 32

Yeah.

Answer 33

Do what is the whole answer?

Answer 34

2x*ln(7+(x^2)^2)…?

Answer 35

You can simplify \((x^2)^2\), can't you?

Answer 36

thank you sooooo much!!!! i got it right :) they took the unsimplified version but it'd be x^4