Question

Find the derivative of the function:
f(x)=4arcsin(x/2)

Answer 1

I did some work and got:

\[\frac{ 4\sqrt{4-x^2} }{ 4-x^2 }\]

Answer 2

\[\Large f(x)\quad=\quad 4\arcsin\left(\frac{x}{2}\right)\]
\[\Large f'(x)\quad=\quad 4\left(\frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^2}}\right)\left(\frac{1}{2}\right)\]

Answer 3

And I guess you did some shinanigans to simplify it down? Hmm ok ok let's see..

Answer 4

\[\Large f'(x)\quad=\quad 2\left(\frac{1}{\sqrt{1-\frac{x^2}{4}}}\right)\]
Multiplying through by 2/2 gives us:\[\Large f'(x)\quad=\quad \frac{4}{\sqrt{4-x^2}}\]

Answer 5

And then I guess you rationalized it, giving usssssss...
Yah looks like you did it correctly! :)

Answer 6

alright perfect.

now if you dont mind helping me with one last question:

find the derivative of g(x)=e^2xarctan(2x)

I got up to:

g'(x)=e^2x\[\frac{ 2 }{ 1+4x^2 }\]

Answer 7

g'(x)=e^2x*2/1+4x^2

I dont know what to do from here

Answer 8

Looks like we'll need to start with product rule, yes? :)\[\Large g(x)=e^2x \arctan(2x)\]
\[\Large g'(x)=\color{royalblue}{(e^{2x})'}\arctan2x+e^{2x}\color{royalblue}{(\arctan2x)'}\]

Answer 9

The blue parts are where we need to take a derivative.
Confusion? :u

Answer 10

I understand. This is what I get:

e^2x(2) * [2/(1+2x)]+e^2x[4/(1+2x)^2]

Answer 11

or is it:

(2) * e^2x * [2/1+4x^2]+ e^2x [16x/(1+4x^2)^2]

Answer 12

ah sorry i ran off to make some food :d mmm let's see...

Answer 13

\[\Large g'(x)\quad =\quad\color{royalblue}{(e^{2x})'}\arctan2x\quad+\quad e^{2x}\color{royalblue}{(\arctan2x)'}\]

\[\Large g'(x)\quad=\quad\color{orangered}{2e^{2x}}\arctan2x\quad+\quad e^{2x}\color{orangered}{\frac{2}{1+(2x)^2}}\]

Answer 14

We should only be differentiating the arctan within the second term, right? :)

Answer 15

That arctan derivative giving you a little trouble?