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Please help with Derivatives - by tonight?
- Thread starter rafeeki92
- Start date
rafeeki92 said: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.
BigGlenntheHeavy
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- Mar 8, 2009
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\(\displaystyle 2) \ Find \ f \ ' \ (x^{2}) \ when \ \frac{d[f(x^{2})]}{dx} \ = \ x^{2}.\)
\(\displaystyle \frac{d[f(x^{2})]}{dx} \ = \ [f \ ' \ (x^{2})](2x) \ = \ x^{2}, \ hence, \ f \ ' \ (x^{2}) \ = \ \frac{x}{2}.\)
\(\displaystyle \frac{d[f(x^{2})]}{dx} \ = \ [f \ ' \ (x^{2})](2x) \ = \ x^{2}, \ hence, \ f \ ' \ (x^{2}) \ = \ \frac{x}{2}.\)
BigGlenntheHeavy
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- Mar 8, 2009
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\(\displaystyle 1) \ Given: \ f[cos(x)] \ = \ x^{2}, \ Find \ \lim_{x\to0}f' [cos(x)]\)
\(\displaystyle Note: \ \frac{d[f[g(x)]]}{dx} \ = \ f'[g(x)]g'(x).\)
\(\displaystyle Hence, \ taking \ the \ derivatives \ of \ both \ sides, \ yields:\)
\(\displaystyle \frac{d[f[cos(x)]]}{dx} \ = \ \frac{d[x^{2}]}{dx}\)
\(\displaystyle f'[cos(x)](-sin(x)) \ = \ 2x, \ f'[cos(x)] \ = \ \frac{-2x}{sin(x)}\)
\(\displaystyle Then, \ \lim_{x\to0}f'[cos(x)] \ = \ \lim_{x\to0}\frac{-2x}{sin(x)} \ = \ -2, \ use \ the \ Marqui.\)
\(\displaystyle Note: \ \frac{d[f[g(x)]]}{dx} \ = \ f'[g(x)]g'(x).\)
\(\displaystyle Hence, \ taking \ the \ derivatives \ of \ both \ sides, \ yields:\)
\(\displaystyle \frac{d[f[cos(x)]]}{dx} \ = \ \frac{d[x^{2}]}{dx}\)
\(\displaystyle f'[cos(x)](-sin(x)) \ = \ 2x, \ f'[cos(x)] \ = \ \frac{-2x}{sin(x)}\)
\(\displaystyle Then, \ \lim_{x\to0}f'[cos(x)] \ = \ \lim_{x\to0}\frac{-2x}{sin(x)} \ = \ -2, \ use \ the \ Marqui.\)