f(x)=sin(1/x), compute f'(-1/x)

Question

$f(x) = \sin\left(\frac{1}{x}\right)$

Compute $f'(-\frac{1}{x})$.

While this problem isn't particularly hard, I noticed desmos gave a different answer from me.

What I did was (to be safe), $f'(x)=-\frac{1}{x^2}\cos\left(\frac{1}{x}\right)$.

Therefore, $f'(-\frac{1}{x})=-\frac{1}{x^4}\cos\left(-x\right)=-\frac{1}{x^4}\cos\left(x\right)$.

Did I make a mistake in here? And if so where?

Checking back to the differentials, df/dv = df/dx * dx/dv. dx/dv = x^2, and df/dx = -1/x^2 cos(1/x). did i make a mistake somewhere in there, is that why desmos shows -cosx as the answer?

what is the correct answer?

score 2 · Answer 1 · answered Mar 30 '17 at 06:41

2

Write $f'(y) = \frac{-1}{y^{2}}\cos (\frac{1}{y})$. Now if $y := -1/x$ then $$ f'(y) = f'(-1/x) = -x^{2}\cos (-x) = -x^{2}\cos (x). $$

answered Mar 30 '17 at 06:41

Yes

20,719

what about chain rule?? – Saketh Malyala Mar 30 '17 at 06:53
Let $g(x):=f(-1/x)$. Then $g'(x) \ne f'(-1/x)$ ! – Fred Mar 30 '17 at 06:56
@Saketh, I dealt with not the chain rule; instead, I was implying that the OP was confused by the symbols used. – Yes Mar 30 '17 at 07:56
1

Wait a minute, I think I understand.
The chain rule says d/dx f(g(x)) = f'(g(x))g'(x)

however, f'(g(x)) never calls for the chain rule.

also, i forgot to substitute -1/x for 1/x^2 when plugging in
– Saketh Malyala Mar 31 '17 at 02:42
That is it! :) ) – Yes Mar 31 '17 at 03:34

score 2 · Accepted Answer · answered Mar 30 '17 at 06:41

2

Your derivative is wrong. $f'(x)=-\frac{1}{x^2}\cos\left(\frac{1}{x}\right)$ is correct.

Hence $f'(-\frac{1}{x})=-\frac{1}{(-\frac{1}{x})^2}\cos\left(\frac{1}{-\frac{1}{x}}\right)=-x^2 \cos(-x)=-x^2 \cos(x)$

answered Mar 30 '17 at 06:41

Fred

77,394

that's what DESMOS got!! – Saketh Malyala Mar 30 '17 at 06:51

f(x)=sin(1/x), compute f'(-1/x)

2 Answers2