Derivative of Sine Function Raised to An Exponent

1. Jun 12, 2013

Justabeginner

1. The problem statement, all variables and given/known data
Find the derivative of (sin x) ^ ((sin(sin x)))

2. Relevant equations

3. The attempt at a solution
I get sin(sin x) * [(sin x) ^ {(sin(sinx)-1)}* cos x] The cos x isn't part of the exponent

Is this right? Thanks :)

2. Jun 12, 2013

haruspex

No, I think it' more complicated than that. First, I would substitute y = sin(x), and use df/dx = df/dy dy/dx. The tricky part is df/dy. For that, try rewriting f(y) as e^(ln f(y)).

3. Jun 12, 2013

Justabeginner

This is what I have:

u= sin x
du= cos x
f(u) = u ^(sin u) du
f'(u)= u^((sin u)-1)* (sin (u) + ((u)(ln u)(cos u)))
And then just substitute sin x in for u?

4. Jun 13, 2013

haruspex

Almost. You have computed df/du ok, but you need df/dx. You mention du = cos(x) (properly, du = cos(x)dx), so maybe you hadn't really forgotten that.

5. Jun 13, 2013

I would use logarithmic differentiation. If $f(x)= (sin(x))^(sin(sin(x))[itex], then [itex]ln(f(x))= sin(sin(x))ln(sin(x)). Now use the product rule and chain rule to find f'(x)/f(x). 6. Jun 13, 2013 darkxponent Using the logrithmich method would make the problem easier. Try it, as suggested by one more post 7. Jun 13, 2013 Justabeginner I'm really confused as to what df/dx would be. If u= sin x, do I rearrange to get the value of x? 8. Jun 14, 2013 darkxponent Hint: df/dx = (df/du)*(du/dx). This is called chain rule. You better use that lograthmic method, it would be easier. Take 9. Jun 14, 2013 haruspex In post #3 you can see that Justabeginner has already cracked that part to obtain df/du. All that remains is to use the chain rule, i.e. multiply by du/dx. 10. Jun 14, 2013 HallsofIvy No, he hasn't. What he had in post #3 was completely wrong. In particular, the derivative of [itex]u^{f(u)}$ is NOT $u^{f(u)-1}f'(u)$

Instead, taking log of both sides of $y= u^{f(u)}$ gives $ln(y)= f(u)ln(u)$ and differentiating both sides of that, with respect to u, $\frac{1}{y}\frac{dy}{du}= f'(u) ln(u)+ \frac{f(u)}{u}$, $\frac{1}{u^{f(u)}}\frac{dy}{du}$$= f'(u)ln(u)+ \frac{f(u)}{u}$, $\frac{dy}{du}= u^{f(u)}(f'(u)ln(u)+ \frac{f(u)}{u}$

Last edited by a moderator: Jun 14, 2013
11. Jun 14, 2013

haruspex

That isn't what he did.
which he wrote as
$\frac{dy}{du}= u^{f(u)-1}(uf'(u)ln(u)+ f(u))$

12. Jun 15, 2013

Justabeginner

This is the train of thought that I was using as I solved for f'(u).
And to finish off I'd just multiply f'(u) by du?

13. Jun 15, 2013

haruspex

By du/dx. Maybe that's what you meant.

14. Jun 16, 2013

Justabeginner

Yes sir, that's what I meant. Thank you very much :)