How to take log on this expression

Question

I am solving exact differential equation, but I am stuck on the step on how to simplify this term or how to rewrite it.

$e^{-2\ln{\sin{x}}}$

yes it is the power of e – Syed Qasim Ahmed Mar 14 '16 at 07:04 — Syed Qasim Ahmed, Mar 14 '16 at 07:04
I want to simplify it – Syed Qasim Ahmed Mar 14 '16 at 07:04 — Syed Qasim Ahmed, Mar 14 '16 at 07:04
I want to simplify it before differentiating – Syed Qasim Ahmed Mar 14 '16 at 07:05 — Syed Qasim Ahmed, Mar 14 '16 at 07:05
Is it this equation?$$e^{-2ln(\sin (x))}$$ – Itakura Mar 14 '16 at 07:07 — Itakura, Mar 14 '16 at 07:07

score 3 · Answer 1 · answered Mar 14 '16 at 07:09

3

So $e^{-2ln(sinx)}=(\sin(x))^{-2ln(e)}=\sin^{-2}(x)$ which can be easily differentiated by chain rule . or division rule.

answered Mar 14 '16 at 07:09

Archis Welankar

15,910

score 3 · Answer 2 · answered Mar 14 '16 at 07:11

3

$$e^{-2 \ln (\sin (x))}=(\sin (x))^{-2}=\csc^2(x)$$

answered Mar 14 '16 at 07:11

Itakura

588

score 3 · Accepted Answer · edited Mar 14 '16 at 07:36

3

Bring the $-2$ inside the logarithm $$e^{\ln(\sin(x)^{-2})} $$ The logarithm and the exponential function cancel leaving $$\sin(x)^{-2}$$ Now you can take the derivative using the chain rule: $$(\sin(x)^{-2})' = -2\sin(x)^{-3} (\sin(x))' = -2\sin(x)^{-3} \cos(x) $$ Which can be rewritten as $$\dfrac{-2\cos(x)}{\sin(x)^3}$$

edited Mar 14 '16 at 07:36

Aritra Das

3,528

answered Mar 14 '16 at 07:13

Jens Renders

4,334

thanks @jens Render Now I think it is the perfect answer. – Syed Qasim Ahmed Mar 14 '16 at 07:15

How to take log on this expression

3 Answers3