SOLUTION: Please help me solve this equation: a) Differentiate {{{ f(x) = cos^(-1) (e^x)/ root(3, x) }}} DO NOT SIMPLIFY b) Simplify {{{ sin ( tan^(-1) (2x) ) }}}

Algebra ->  Test -> SOLUTION: Please help me solve this equation: a) Differentiate {{{ f(x) = cos^(-1) (e^x)/ root(3, x) }}} DO NOT SIMPLIFY b) Simplify {{{ sin ( tan^(-1) (2x) ) }}}      Log On


   

Question 355601: Please help me solve this equation:
a) Differentiate +f%28x%29+=+cos%5E%28-1%29+%28e%5Ex%29%2F+root%283%2C+x%29+ DO NOT SIMPLIFY

b) Simplify +sin+%28+tan%5E%28-1%29+%282x%29+%29+
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29+=+cos%5E%28-1%29+%28e%5Ex%29%2F+root%283%2C+x%29+
This problem uses the several levels of the chain rule. First we will
Let u = cos%5E%28-1%29+%28e%5Ex%29 and v+=+root%283%2C+x%29+=+x%5E%281%2F3%29.
This makes f = u/v and from this we know that
f' = (v*u' - u*v')/v^2
For this we will need u' and v':
u' = %281%2Fsqrt%281+-+%28e%5Ex%29%5E2%29%29%2A%28d%2Fdx%29%28e%5Ex%29
Since the derivative of e%5Ex is e%5Ex this becomes:
u' = %281%2Fsqrt%281-e%5E%282x%29%29%29%2Ae%5Ex+=+e%5Ex%2Fsqrt%281-e%5E%282x%29%29
and v' = %281%2F3%29x%5E%281-%281%2F3%29%29+=+%281%2F3%29x%5E%28-2%2F3%29
Substituting u, u', v and v' into the f' equation we get:
f' =
And if you're not supposed to simplify, then I guess this mess is an acceptable answer.

b) +sin+%28+tan%5E%28-1%29+%282x%29+%29+
For this one, picture a right triangle. For one of the acute angles we want the tangent to be 2x. In other words we want the ratio of opposite/adjacent to be 2x. An opposite side of 2x and an adjacent side of 1 would give us this ratio. We want to find the sin of this angle. Since sin is opposite/hypotenuse, we will need an expression for the hypotenuse. For this we can use the Pythagorean Theorem:
Opposite^2 + Adjacent^2 = Hypotenuse^2
Putting our expressions for the opposite side and adjacent side into this we get:
%282x%29%5E2+%2B+%281%29%5E2 = Hypotenuse^2
which simplifies to:
4x%5E2+%2B+1 = Hypotenuse^2
So the Hypotenuse is sqrt%284x%5E2+%2B+1%29
Now we can express the sin ratio:
+sin+%28+tan%5E%28-1%29+%282x%29+%29+=+%282x%29%2Fsqrt%284x%5E2+%2B+1%29