Normal integral formula problem

[tee yeh hun]

Homework Statement

Given that f'(x)=(cosx)^ (1/3) then find out dy/dx f(ln cosx)=?

The answer given is -(cos(lncos)) ^ (1/3) tanx

Anyone can solve it? I tried and got lost with it

Homework Equations

Using the normal integral formula and derieve it again

The Attempt at a Solution

 

[hunt_mat]

By the chain rule we have:
$$f(x)=\cos^{\frac{1}{3}}x=u^{\frac{1}{3}}$$
Use the chain rule to obtain
$$\frac{df}{dx}=\frac{df}{du}\frac{du}{dx}$$

 

[HallsofIvy]

It doesn't really take a genius, just perseverence and steady, step by step work.

By the chain rule, the derivative of f(ln(cos(x))) is f'(ln(cox(x))) times the derivative of ln(cos(x)). And that is the 1/cos(x) times the derivative of cos(x) which is -sin(x).

The derivative of f(ln(cos(x))) is f'(ln(cos(x))) times -sin(x)/cos(x).

Since $f'(x)= (cos(x))^{1/3}$, [/itex]f'(ln(cos(x))= (cos(ln(cos(x)))))^{1/3}[/itex].

Putting those together gives exactly what your book says.

 

[tee yeh hun]

here are my further works

I found out my problem it can't be done

The problem is i can't increase the power of cos x to (cos x)^2 so there is only one shortest way to solve it

 

[tee yeh hun]

Here is the shortest way

f'(x )=cos(x )^(1/3)
f'(ln(cos(x )))=cos(ln(cos(x )))^1/3* d(ln(cos(x )))/dy* d(cos(x ))/dy
=cos(ln(cos(x )))^1/3 * 1/cos( x) * (-sin(x ))
=cos(ln(cos(x )))^1/3 * -sin(x )/cos(x )
=cos(ln(cos(x )))^1/3 * -tan(x )

here is the shortest way to finish it

done by anexgenesis

 

[hunt_mat]

If
$$u=\cos x$$
and
$$f(u)=u^{\frac{1}{3}}$$
then
$$df/du=u^{-\frac{2}{3}}/3$$
that is your problem.

 

[vela]

If
$$u=\cos x$$
and
$$f(u)=u^{\frac{1}{3}}$$
then
$$df/du=u^{-\frac{2}{3}}/3$$
that is your problem.

You're misreading the problem. He's given df/dx=(cos x)^1/3, not f(x).

 

[hunt_mat]

Oh right. I didn't see the dash!

 

[tee yeh hun]

Problem solved :)