Chain/Composite rule

[hoplessatmaths]

Hi can someone please check my work?
Firsly I have to use the composite rule to differentiate the function:
f(x)=exp(2/3sinx) [exp(x) =e^x]

My answer:
f'(x)=e^x(2/3sin(x) (2/3sin(x))
f'(x)=e^x(2/3sin(x)) 2/3cos(x)
f'(x)=2/3cos(x) e^x(2/3sin(x)

Secondly I have to use the Product rule and he answer above to show that the function g(x)=cos x e^x(2/3 sin x) where (0_< x _<2pi) has the derivative: g'(x) =1/3 (2-3 sin x -2 sin^2x)e^x(2/3sin x)

My answer:
cos(x)2/3 cos(x)e^x(2/3sin(x)+(-sin(x)e^x(2/3sin(x))
2/3cos^2(x)e^x(2/3sin(x))e^x(2/3sin(x))
g'(x)=(2/3cos^2(x))-sin(x)) e^x(2/3sin(x))

 

[galactus]

Looks good. Except be careful of that extra x you have in the first one.

Should be \(\displaystyle \frac{2}{3}cos(x)\cdot e^{\frac{2}{3}sin(x)}\)

The last one is \(\displaystyle e^{\frac{2}{3}sin(x)}\left(\frac{2}{3}cos^{2}(x)-sin(x)\right)\)

Which I believe is what you have. Good work.