Question

Can someone just check my work?

Find the derivitive of the given function:

2sin^-1(5x^4): I got 40x^3/sqrt(1-(5x^4)^2

tan^-1sqrt(5x) I got 5/(20x^2+2)sqrt(tan^-1*5x)

Answer 1

Mmmm first one looks good. You applied chain rule. good good.

Answer 2

\[\Large\rm \frac{d}{dx}\arctan\left(\color{orangered}{\sqrt{5x}}\right)=\frac{1}{1+\left(\color{orangered}{\sqrt{5x}}\right)^2}\cdot \frac{1}{2\sqrt{5x}}\cdot 5\]Here is how the other problem would work out.
First part is the arctan derivative.
Then the second is from the chain rule, derivative of square root is 1 over 2 square roots, yes?

Then the last part is from the 5x derivative.

Answer 3

wow that's complicated so would it be 5/sqrtx *(10x+2) ?

Answer 4

So you distributed the 2 and combined all the fractions?
We should still have a sqrt5 somewhere in the denominator.
Or you can simplify a little further since you have a 5 on top.\[\Large\rm \frac{\sqrt5}{\sqrt x(10x+2)}\]

Answer 5

Thanks so much! do you mind looking over one more really quick thing?
it's Determine the values of x for which the function is differentiable.
y=sqrt(x^2+9) I got all values

Answer 6

or is it -3,3,0?

Answer 7

Ummm I think what we want to do is..

Answer 8

first deriv and set = to 0?

Answer 9

\[\Large\rm f(x)=\sqrt{x^2-9},\qquad\qquad\to\qquad\qquad f'(x)=\frac{x}{\sqrt{x^2-9}}\]Yah looks good, set the first derivative equal to zero.
Then find out where the first derivative is `undefined`.

Answer 10

it was +9

Answer 11

Oh woops :\

Answer 12

so all values?

Answer 13

Mmmmmm yah I think so +_+
I don't see any troublesome spots.

Answer 14

Thanks!