Differentiation.

Hi Max.

It depends where the brackets go.  I know there is a convention that I always get confused so I am probably not answering the question that you have asked.    :)   

$f(x)=\left(e^{sin(x)}\right)^{cos(x)}\\ f(x)=e^{cos(x)sin(x)}\\ f(x)=e^{0.5*{sin(2x)}}\\ f'(x)=[cos(2x)]e^{0.5{sin(2x)}}$
 

Now I think that guest has answered this question.  It is harder....   

$f(x)=e^{\left(sin(x)^{cos(x)}\right)}\\ $

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx((e^(sin(x)))^(cos(x)))
The derivative of f(x) is f'(x):
f'(x) = d/dx((e^(sin(x)))^(cos(x)))
Express (e^(sin(x)))^(cos(x)) as a power of e: (e^(sin(x)))^(cos(x)) = e^(log((e^(sin(x)))^(cos(x)))) = e^(cos(x) log(e^(sin(x)))):
f'(x) = d/dx(e^(cos(x) log(e^(sin(x)))))
Using the chain rule, d/dx(e^(cos(x) log(e^(sin(x))))) = ( d e^u)/( du) 0, where u = cos(x) log(e^(sin(x))) and ( d)/( du)(e^u) = e^u:
f'(x) = d/dx(cos(x) log(e^(sin(x)))) e^(cos(x) log(e^(sin(x))))
Express e^(cos(x) log(e^(sin(x)))) as a power of e^(sin(x)): e^(cos(x) log(e^(sin(x)))) = e^(log((e^(sin(x)))^(cos(x)))) = (e^(sin(x)))^(cos(x)):
f'(x) = (e^(sin(x)))^(cos(x)) d/dx(cos(x) log(e^(sin(x))))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cos(x) and v = log(e^(sin(x))):
f'(x) = log(e^(sin(x))) d/dx(cos(x))+cos(x) d/dx(log(e^(sin(x)))) (e^(sin(x)))^(cos(x))
The derivative of cos(x) is -sin(x):
f'(x) = (e^(sin(x)))^(cos(x)) (cos(x) (d/dx(log(e^(sin(x)))))+-sin(x) log(e^(sin(x))))
Simplify log(e^(sin(x))) using the identity log(a^b) = b log(a):
f'(x) = (e^(sin(x)))^(cos(x)) (-log(e^(sin(x))) sin(x)+d/dx(sin(x) log(e)) cos(x))
The derivative of sin(x) is cos(x):
f'(x) = (e^(sin(x)))^(cos(x)) (-log(e^(sin(x))) sin(x)+cos(x) cos(x))
Simplify the expression:
f'(x) = (e^(sin(x)))^(cos(x)) (cos^2(x)-log(e^(sin(x))) sin(x))
Expand the left hand side:
Answer: |f'(x) = (e^(sin(x)))^(cos(x)) (cos^2(x)-log(e^(sin(x))) sin(x))

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(2 e^(csc(x)) x)
The derivative of f(x) is f'(x):
f'(x) = d/dx(2 e^(csc(x)) x)
Factor out constants:
f'(x) = 2 d/dx(e^(csc(x)) x)
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = e^(csc(x)) and v = x:
f'(x) = 2 x d/dx(e^(csc(x)))+e^(csc(x)) d/dx(x)
Using the chain rule, d/dx(e^(csc(x))) = ( d e^u)/( du) 0, where u = csc(x) and ( d)/( du)(e^u) = e^u:
f'(x) = 2 (e^(csc(x)) (d/dx(x))+e^(csc(x)) d/dx(csc(x)) x)
The derivative of csc(x) is -cot(x) csc(x):
f'(x) = 2 (e^(csc(x)) (d/dx(x))+-cot(x) csc(x) e^(csc(x)) x)
The derivative of x is 1:
f'(x) = 2 (-e^(csc(x)) x cot(x) csc(x)+1 e^(csc(x)))
Simplify the expression:
f'(x) = 2 (e^(csc(x))-e^(csc(x)) x cot(x) csc(x))
Expand the left hand side:
Answer: |f'(x) = 2 (e^(csc(x))-e^(csc(x)) x cot(x) csc(x))
                                           

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(cot(x) (2^(2 x)+(log(x))/(log(2))))
The derivative of f(x) is f'(x):
f'(x) = d/dx(cot(x) (2^(2 x)+(log(x))/(log(2))))
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cot(x) and v = 2^(2 x)+(log(x))/(log(2)):
f'(x) = (2^(2 x)+(log(x))/(log(2))) d/dx(cot(x))+cot(x) d/dx(2^(2 x)+(log(x))/(log(2)))
The derivative of cot(x) is -csc^2(x):
f'(x) = cot(x) (d/dx(2^(2 x)+(log(x))/(log(2))))+(2^(2 x)+(log(x))/(log(2))) -csc(x)^2
Differentiate the sum term by term and factor out constants:
f'(x) = -csc^2(x) (2^(2 x)+(log(x))/(log(2)))+d/dx(2^(2 x))+(d/dx(log(x)))/(log(2)) cot(x)
Using the chain rule, d/dx(2^(2 x)) = ( d2^u)/( du) 0, where u = 2 x and ( d)/( du)(2^u) = 2^u log(2):
f'(x) = -csc^2(x) (2^(2 x)+(log(x))/(log(2)))+cot(x) ((d/dx(log(x)))/(log(2))+2^(2 x) log(2) d/dx(2 x))
Factor out constants:
f'(x) = -csc^2(x) (2^(2 x)+(log(x))/(log(2)))+cot(x) ((d/dx(log(x)))/(log(2))+2 d/dx(x) 2^(2 x) log(2))
Simplify the expression:
f'(x) = -csc^2(x) (2^(2 x)+(log(x))/(log(2)))+cot(x) (2^(1+2 x) log(2) (d/dx(x))+(d/dx(log(x)))/(log(2)))
The derivative of x is 1:
f'(x) = -csc^2(x) (2^(2 x)+(log(x))/(log(2)))+cot(x) ((d/dx(log(x)))/(log(2))+1 2^(1+2 x) log(2))
The derivative of log(x) is 1/x:
f'(x) = -csc^2(x) (2^(2 x)+(log(x))/(log(2)))+cot(x) (2^(1+2 x) log(2)+1/x/(log(2)))
Expand the left hand side:
Answer: |f'(x) = cot(x) (1/(x log(2))+2^(1+2 x) log(2))-csc^2(x) (2^(2 x)+(log(x))/(log(2)))

It was soooo messy!! Could anyone please use LaTeX like I did?

Second one :D

(2^x)(e^{\frac{1}{\sin x}})

$f(x)=(2^x)(e^{\frac{1}{sin x}})\\ f(x)=(2^x)e^{(sinx)^{-1}}\\ f(x)=uv\;\;\;where\;\;\;u=2^x\;\;\;and\;\;\;v=e^{(sinx)^{-1}}\\~\\ v=e^{(sinx)^{-1}}\\ \frac{dv}{dx}=-(sinx)^{-2}(cosx)e^{(sinx)^{-1}}\\ \frac{dv}{dx}=\frac{-(cosx)e^{(sinx)^{-1}}}{(sinx)^{2}}\\~\\ u=2^x\\ ln(u)=ln(2^x)\\ ln(u)=xln(2)\\ x=\frac{ln(u)}{ln2}\\ \frac{dx}{du}=\frac{1}{u*ln2}\\ \frac{du}{dx}=u*ln2\\ \frac{du}{dx}=2^x*ln2\\ $

$f(x)=(2^x)(e^{\frac{1}{sin x}})\\ f(x)=(2^x)e^{(sinx)^{-1}}\\ f(x)=uv\;\;\;where\;\;\;u=2^x\;\;\;and\;\;\;v=e^{(sinx)^{-1}}\\~\\ v=e^{(sinx)^{-1}}\\ \frac{dv}{dx}=\frac{-(cosx)e^{(sinx)^{-1}}}{(sinx)^{2}}\\~\\ u=2^x\\ \frac{du}{dx}=2^x*ln2\\ \mbox{Using product rule}\\ f'(x)=2^x*\;\frac{-(cosx)e^{(sinx)^{-1}}}{(sinx)^{2}}\;\;+\;\;e^{(sinx)^{-1}}\;*\;2^xln2\\ f'(x)=2^x*\;\frac{-(cosx)e^{cosec(x)}}{(sinx)^{2}}\;\;+\;\;e^{cosec(x)}\;*\;2^xln2\\ f'(x)=2^xe^{cosec(x)}\left(\;\frac{-(cosx)}{(sinx)^{2}}\;\;+\;\;ln2\right)\\ f'(x)=\frac{2^xe^{cosec(x)}}{sin^2x}\;(ln2\;sin^2(x)\;-\;cosx)\\ $

My answer is the same as Wolfram Alpha although it is arranced a little differently     

http://www.wolframalpha.com/input/?i=differentiate++2%5Ex*e%5E(1%2Fsinx)