sin^-1(5x+4)....I got (5)/(sqrt1-x^2)...not sure where i went wrong

Question

anonymous · Answer

Are you simplifying? Or are you taking the derivative?

anonymous · Answer

Ok if you're taking the derivative you should follow the formula
$$d(\sin^{-1} u)/dx=du/(\sqrt{1-u^2}dx)$$
So the derivative would look similar to your answer,
You first take the derivative of u (5x+4) and you get 5 as the numerator like you did.
Next, the denominator has to be $$\sqrt{1-(5x+4)^2}=\sqrt{-25x^2-40x-15}$$
So your answer would be
$$5/\sqrt{-25x^2-40x-15}$$

anonymous · Answer

ok cuz i have no idea where i got the denmontor wrong

anonymous · Answer

What do you mean? You're equation is right, you just have to use everything that's inside the sin-1 instead of just x when you square it. So in the denominator you should put 5x+4 instead of just the x.

anonymous · Answer

but im now really solveing for x though...

anonymous · Answer

Right, but you're taking the derivative, so the denominator should have (5x+4)^2 instead of x.

anonymous · Answer

mmm i guess.....im just stressing out over the chain rule and derivtive