1

I have this function $\frac{\sin^2 x}{1-\cos x}$.

$\frac{\sin^2x}{1-\cos x}=\frac{1-\cos^2x}{1-\cos x}=1+\cos x\;$. Thus the derivative of $1 + \cos x\; = -\sin x\;$.

However by, L'Hopital's rule, I obtain

$\frac{\sin^2x}{1-\cos x}=\frac{2\sin x\cos x}{\sin x}=2\cos x\;$. Thus the derivative of $1 + \cos x\; = 2 \cos x\;$.

What's wrong with this?

1 Answers1

2

We use L'Hopital's rule for evaluating certain limits, not for evaluating derivatives! Your initial answer is correct, since L'Hopital is not at all appropriate here.

Note: If your function did not simplify so nicely, you could have used the quotient rule in order to find the derivative, since it is of a function that is the quotient of two other functions for which derivatives exist.

If the function one wishes to differentiate, f(x), can be written as $$f(x) = \frac{g(x)}{h(x)}$$ then the rule states that the derivative of $\dfrac{g(x)}{h(x)}$ is

$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.$$

kahen
  • 15,760
amWhy
  • 209,954
  • Thanks, @kahen, for the edit of my stupid grammatical error! :-X – amWhy Sep 21 '13 at 15:33
  • Thanks, wow - I'm soo rusty in my calculus. I have a lot of brushing up to do! –  Sep 21 '13 at 15:37
  • You're welcome. It'll "come back" quicker than it took the first time through! – amWhy Sep 21 '13 at 15:40
  • Yes definitely. The key is to rely on "learning savings" 'cuz forgetting is better than not knowing! –  Sep 21 '13 at 15:41