Hi can some please check my working out and answer?
Use the composite rule to differentiate the function
f(x)=e^x(2/3sin x)
My answer:
e^x(2/3sinx)
2/3 sin x = 2sin x/3
e^x(2sin x/3)
2e^xsin x/3
Thanks in advance
It would help if you used more parentheses. The "composite" rule, I would think of as involving composite functions: f(g(x)) but what you have written appears to be (e^x) ((2/3)sin(x)) rather than the composite e^((2/3)x sin(x)). In any case, you don't seem to differentiated at all! You final "answer", 2(e^x)(sin(x))/3, is just the original function. What is the derivative of that function?
This now says $\displaystyle f(x)= e^x (\frac{2}{3}sin(x))$ is that what you intended?
I have no idea what you mean by this! it is certainly NOT the function above. Did you mean that as f'(x)? It is not for either $\displaystyle f(x)= e^x(\frac{2}{3}sin(x))$ or $\displaystyle f(x)= e^{\frac{2}{3}x sin(x)}$.f(x)=e^x(2/3sin(x) (2/3sin(x))
If you meant $\displaystyle f(x)= e^{\frac{2}{3}x sin(x)}$, its derivative is itself, $\displaystyle e^{\frac{2}{3}x sin(x)}$ times the derivative of $\displaystyle \frac{2}{3}x sin(x)$ which is NOT $\displaystyle \frac{2}{3}cos(x)$.=e^x(2/3sin(x) 2/3cos(x)
=2/3cos(x) e^x(2/3sin(x))
I'm stuck on trying to differentiate f(x)=e^x(2/3sin x) as well and I've tangled myself up into a complete mind block! I think it's the 2/3 aspect confusing me the most, I've never been quite sure what to do with fractions when they pop up like this in functions. D'oh. Any help/advice would be much appreciated!
If this really is $\displaystyle f(x)= e^x((2/3)sin(x))$, then the "2/3", a constant, just multiplies the whole thing. The derivative is $\displaystyle f'(x)= (2/3)(e^x sin(x)+ e^x cos(x))= (2/3)e^x(sin(x)+ cos(x))$
If it is actually the composite function, as suggested by the title, $\displaystyle f(x)= e^{(2/3)x sin(x)}$ the, by the composite rule (also called the "chain rule") $\displaystyle f'(x)= e^{(2/3)x sin(x)}\left((2/3)x sin(x)\right)'= 2/3(sin(x)+ xcos(x))e^{(2/3)x sin(x)}$
the rule is $\displaystyle \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx}$ , where $\displaystyle u$ is a function of $\displaystyle x$.
for your problem ...
$\displaystyle y = e^{\frac{2}{3}\sin{x}}$
$\displaystyle \frac{dy}{dx} = e^{\frac{2}{3}\sin{x}} \cdot \frac{2}{3}\cos{x}$