Derivative of Sine Function Raised to An Exponent

[Justabeginner]

Homework Statement

Find the derivative of (sin x) ^ ((sin(sin x)))

Homework Equations

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 :)

 

[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)).

 

[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?

 

[haruspex]

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?

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.

 

[HallsofIvy]

I would use logarithmic differentiation. If [itex]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).

 

[darkxponent]

Homework Statement

Find the derivative of (sin x) ^ ((sin(sin x)))

Homework Equations

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 :)

Using the logrithmich method would make the problem easier. Try it, as suggested by one more post

 

[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?

 

[darkxponent]

Hint: df/dx = (df/du)*(du/dx). This is called chain rule.

You better use that lograthmic method, it would be easier. Take

 

[haruspex]

You better use that lograthmic method,

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.

 

[HallsofIvy]

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.

No, he hasn't. What he had in post #3 was completely wrong.

In particular, the derivative of $u^{f(u)}$ 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}$, $\frac{dy}{du}= u^{f(u)}(f'(u)ln(u)+ \frac{f(u)}{u}$

 

[haruspex]

No, he hasn't. What he had in post #3 was completely wrong.

In particular, the derivative of $u^{f(u)}$ is NOT $u^{f(u)-1}f'(u)$

That isn't what he did.

$\frac{dy}{du}= u^{f(u)}(f'(u)ln(u)+ \frac{f(u)}{u}$

which he wrote as
$\frac{dy}{du}= u^{f(u)-1}(uf'(u)ln(u)+ f(u))$

 

[Justabeginner]

That isn't what he did.

which he wrote as
$\frac{dy}{du}= u^{f(u)-1}(uf'(u)ln(u)+ f(u))$

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?

 

[haruspex]

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?

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

 

[Justabeginner]

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