1. ## Composite Rule

Use the composite rule to differentiate the function
f(x)=e^x(2/3sin x)

e^x(2/3sinx)

2/3 sin x = 2sin x/3

e^x(2sin x/3)

2e^xsin x/3

2. Well it's not clear what the expression is. Use LaTeX. If the expression is what I think it is, then you haven't done anything at all to it.

3. Originally Posted by stewpot

Use the composite rule to differentiate the function
f(x)=e^x(2/3sin x)

e^x(2/3sinx)

2/3 sin x = 2sin x/3

e^x(2sin x/3)

2e^xsin x/3

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?

4. Originally Posted by HallsofIvy
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?
f(x)=e^x(2/3sin x)

f(x)=e^x(2/3sin(x) (2/3sin(x))

=e^x(2/3sin(x) 2/3cos(x)

=2/3cos(x) e^x(2/3sin(x))

5. Originally Posted by stewpot
f(x)=e^x(2/3sin x)
This now says $\displaystyle f(x)= e^x (\frac{2}{3}sin(x))$ is that what you intended?

f(x)=e^x(2/3sin(x) (2/3sin(x))
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 $\displaystyle f(x)= e^{\frac{2}{3}x sin(x)}$.

=e^x(2/3sin(x) 2/3cos(x)

=2/3cos(x) e^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 $\displaystyle \frac{2}{3}cos(x)$.

6. ## Got the same problem!

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!

7. 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") $\displaystyle f'(x)= e^{(2/3)x sin(x)}\left((2/3)x sin(x)\right)'= 2/3(sin(x)+ xcos(x))e^{(2/3)x sin(x)}$

8. ## composite rule

I am still unsure as to whether this formula has been entered correctly. The correct formula is in the attached file.

Can you point me in the right direction please?

9. the rule is $\displaystyle \frac{d}{dx} e^u = e^u \cdot \frac{du}{dx}$ , where $\displaystyle u$ is a function of $\displaystyle x$.

$\displaystyle y = e^{\frac{2}{3}\sin{x}}$
$\displaystyle \frac{dy}{dx} = e^{\frac{2}{3}\sin{x}} \cdot \frac{2}{3}\cos{x}$