Question

Differentiate

Answer 1

\[\Huge y=\sin^{-1}(\frac{5x+12\sqrt{1-x^2}}{13})\]

Answer 2

Remember the derivative of arcsine? :)

Answer 3

1/sqrt{1-x^2}

Answer 4

Oh boy, our solution is going to look really ugly for this one lol.

Answer 5

we won't apply the derivative directly

Answer 6

The \(\large x\) that you posted, in your derivative, corresponds to EVERYTHING inside of the arcsine, ok?

So this derivative will give us,\[\large y'=\frac{1}{\sqrt{1-\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)^2}}\color{royalblue}{\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)'}\]

Answer 7

You won't?

Answer 8

lets substitute something and simplify it so it bcomes easy

Answer 9

The blue term shows up due to the chain rule, after taking the derivative of arcsine, we have to multiply by the derivative of the inner function.

Answer 10

I guess you could... hmm

Answer 11

It'll be the exact same process as this, you're just doing the chain rule off to the side.
If that makes it easier for you though, then that's a good idea.

Answer 12

something like x=tantheta

Answer 13

You could do x=sin theta.
That would deal with the square root nicely.

I really really don't think that's going to make the problem easier though.

Answer 14

If you really want to, you can make the substitution,\[\large u=\frac{5x+12\sqrt{1-x^2}}{13}\]

And maybe that will simplify things down a little bit.\[\large (\sin^{-1}u)' \qquad \rightarrow\qquad \frac{1}{\sqrt{1-u^2}}u'\]Then we just need the derivative of u.


Are you suppose to do this one without using the arcsine derivative rule?