Question

Diffentiate the following with respect to x

Answer 1

\[\tan^{-1} (e ^{-((x+1)^2)/4})\]

Answer 2

tan^-1(x)=1/(x^2+1)

Answer 3

the derivative of tan^-1 = that

Answer 4

Yes, @TylerD is correct. \[d(\tan^{-1}(x)) =\frac{1}{1+x^2}\]So if we let \(u =\dfrac{(x+1)^{-2}}{4} \), then: \[d(\tan^{-1}(u))=\frac{1}{1+u^2} \cdot u'\] Find your \(u'\)

Answer 5

@Jhannybean what about the e

Answer 6

Well, since exponents are messly, let's use \(\exp\) instead...\[
\tan^{-1} \left(\exp \left(-\frac{(x+1)^2}4\right)\right)

\]

Answer 7

First we will let \(u = -(x+2)^2/4\): \[
\tan^{-1}(\exp(u))
\]Then let's let \(v = \exp(u)\):\[
\tan^{-1}(v)
\]This we can differentiate easily.\[
d(\tan^{-1}(v)) = \frac{dv}{1+v^2}
\]Now we have to back substitute... \[
d(\tan^{-1}(v)) = \frac{1}{1+(\exp(u))^2}d\left(\exp(u)\right)
\]

Answer 8

Another easy derivative: \[
d(\tan^{-1}(v)) = \frac{1}{1+(\exp(u))^2}d\left(\exp(u)\right) = \frac{1}{1+(\exp(u))^2}\exp(u)~du
\]

Answer 9

We have to keep backsubbing.

Answer 10

It gets a bit messy... \[
d(\tan^{-1}(v)) = \frac{\exp\left(-\frac{(x+2)^2}{4}\right)}{1+\left(\exp\left(-\frac{(x+2)^2}{4}\right)\right)^2}~d\left(-\frac{(x+2)^2}{4}\right)
\]

Answer 11

Let \(y=x+2\) and: \[
d(-y^2/4) = (-2y/4)~dy = (-(x+2)/2)~d(x+2) = (-(x+2)/2)~dx
\]

Answer 12

All together...\[
d\left[\tan^{-1} \left(\exp \left(-\frac{(x+1)^2}4\right)\right)\right]=
- \frac{(x+2)\exp\left(-\frac{(x+2)^2}{4}\right)}{2+2\exp\left(-\frac{(x+2)^2}{2}\right)}dx
\]

Answer 13

\[
dy=\color{red}{\frac{dy}{dx}}dx
\]Byt the way, our derivative is the red part; basically it's the right hand side without the \(dx\).

Answer 14

And like it or not, we can probably simply more, since those \(\exp\) functions follow the rules of exponents.

Answer 15

THANK YOU SO MUCH. it was a lot more than I thought it would be