Question

The derivative of $y=lo{g}_{e}sin\left(e^x\right)$ with respect to x will be

• A. $cosec\left(e^x\right)cos\left(e^x\right).e^x$
• B. $cos\left(e^x\right).e^x$
• C. $\frac{1}{sin\left(e^x\right)}$
• D. $cosec\left(e^x\right)cos\left(e^x\right)$

Answer 1

The correct option is A $cosec\left(e^x\right)cos\left(e^x\right).e^x$

It’s okay if you did not get it. Just Remember that
$\frac{d\left(lo{g}_e\right(x\left)\right)}{dx}=\frac{1}{x}$ only if the argument of
$lo{g}_e$ is x. If not, chain rule has to be applied. Same holds for other functions as well.
Here $y=lo{g}_e\left(sin\right(e^x\left)\right)$
Let’s say $sin\left(e^x\right)=u$
$e^x=v$

$y=lo{g}_{e}u$ But u is a function of $v=e^x$ and v is a function of x
So $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}$ [As per chain rule]

So $\frac{dy}{dx}=\frac{1}{u}\times cosv\times e^x$

$=\frac{1}{sin\left(e^x\right)}\times cos\left(e^x\right)\times e^x=cosec\left(e^x\right)cos\left(e^x\right)\times e^x$