Please help with Derivatives - by tonight?

[rafeeki92]

1) Given f(cosx) = x2, find limx->0 f'(cosx)

For this one, I tried this...but didn't reach any sort of conclusion:
f'(cosx) = limx->cosx [f(x) - x2]/(x-cosx)

2) Find f'(x2) if d/dx[f(x2)] = x2

I wasn't sure how to approach this one at all.

 

[daon]

1) Given f(cosx) = x2, find limx->0 f'(cosx)

For this one, I tried this...but didn't reach any sort of conclusion:
f'(cosx) = limx->cosx [f(x) - x2]/(x-cosx)

2) Find f'(x2) if d/dx[f(x2)] = x2

I wasn't sure how to approach this one at all.

1) Youre supposed to evaluate the limit as x-> 0 f'(cosx). Using the chain rule we see f(cos(x)) = x^2 => f'(cosx)sin(x) = 2x => f'(cos(x)) = 2*(x/sinx) Then take the limit of both sides.

2) Use a similar approach to the above.

 

[rafeeki92]

Oh thank you so much. I'm not sure how I should approach the second question, though. Any hints?

 

[BigGlenntheHeavy]

$\displaystyle 2) \ Find \ f \ ' \ (x^{2}) \ when \ \frac{d[f(x^{2})]}{dx} \ = \ x^{2}.$

$\displaystyle \frac{d[f(x^{2})]}{dx} \ = \ [f \ ' \ (x^{2})](2x) \ = \ x^{2}, \ hence, \ f \ ' \ (x^{2}) \ = \ \frac{x}{2}.$

 

[BigGlenntheHeavy]

$\displaystyle 1) \ Given: \ f[cos(x)] \ = \ x^{2}, \ Find \ \lim_{x\to0}f' [cos(x)]$

$\displaystyle Note: \ \frac{d[f[g(x)]]}{dx} \ = \ f'[g(x)]g'(x).$

$\displaystyle Hence, \ taking \ the \ derivatives \ of \ both \ sides, \ yields:$

$\displaystyle \frac{d[f[cos(x)]]}{dx} \ = \ \frac{d[x^{2}]}{dx}$

$\displaystyle f'[cos(x)](-sin(x)) \ = \ 2x, \ f'[cos(x)] \ = \ \frac{-2x}{sin(x)}$

$\displaystyle Then, \ \lim_{x\to0}f'[cos(x)] \ = \ \lim_{x\to0}\frac{-2x}{sin(x)} \ = \ -2, \ use \ the \ Marqui.$