Question 355601
{{{ f(x) = cos^(-1) (e^x)/ root(3, x) }}}
This problem uses the several levels of the chain rule. First we will
Let u = {{{cos^(-1) (e^x)}}} and {{{v = root(3, x) = x^(1/3)}}}.
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' = {{{(1/sqrt(1 - (e^x)^2))*(d/dx)(e^x)}}}
Since the derivative of {{{e^x}}} is {{{e^x}}} this becomes:
u' = {{{(1/sqrt(1-e^(2x)))*e^x = e^x/sqrt(1-e^(2x))}}}
and v' = {{{(1/3)x^(1-(1/3)) = (1/3)x^(-2/3)}}}
Substituting u, u', v and v' into the f' equation we get:
f' = {{{((x^(1/3))*(e^x/sqrt(1-e^(2x))) - (cos^(-1) (e^x))*((1/3)x^(-2/3)))/(x^(1/3))^2}}}
And if you're not supposed to simplify, then I guess this mess is an acceptable answer.<br>
b) {{{ sin ( tan^(-1) (2x) ) }}}
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:
{{{(2x)^2 + (1)^2}}} = Hypotenuse^2
which simplifies to:
{{{4x^2 + 1}}} = Hypotenuse^2
So the Hypotenuse is {{{sqrt(4x^2 + 1)}}}
Now we can express the sin ratio:
{{{ sin ( tan^(-1) (2x) ) = (2x)/sqrt(4x^2 + 1)}}}