# Where have I gone wrong in finding the derivative of $2^{\sin x}$?

I was finding the derivative of $2^{\sin x}$.

My attempt $(1)$-

$$y=2^{\sin x}$$

$$\implies\ln y=\ln2^{\sin x}$$

$$\implies\log_ey=\ln2^{\sin x}$$

$$\displaystyle\implies e^{\ln2^{\sin x}}=y$$

$$\implies\frac{d}{dx}y=\frac{d}{dx}e^{\ln 2^{\sin x}}$$

Now,I know $\frac{d}{dx}e^x=e^x$ but what about $\frac{d}{dx}e^{\ln 2^{\sin x}}$.

How do I simplify it?

My attempt $(2)$-

$$2^{\sin x}=y$$

$$\log_2y=\sin x$$

$$\frac{d}{dx}\log_2y=\frac{d}{dx}\sin x$$

Now,I feel that chain rule can be used some how in the LHS to get $\frac{dy}{dx}$ which is our required derivative,but I am having trouble applying chain rule.

What to do?

Thanks a lot for any help!!

• What you should have done is note that $\ln 2^{\sin x} = \ln 2\cdot\sin x$. Jun 30 '16 at 3:14
• The first four lines do not help; they are like saying $f(x)=f(x)+3-3$. Jun 30 '16 at 3:17
• @CameronWilliams How did you get that? Jun 30 '16 at 3:19
• @tatan From the properties of logarithms. Jun 30 '16 at 3:21
• @CameronWilliams That was trivial...missed it out...thanks by the way... Jun 30 '16 at 3:22

I feel like it will be easier to go off of Attempt 1, so I'm going to do that: $$\frac{dy}{dx}=\frac{d}{dx}e^{\ln 2^{\sin x}}$$ Now, we know that $\ln2^{\sin x}=\sin x\ln 2$, so make the substitution: $$\frac{dy}{dx}=\frac{d}{dx}e^{\sin x\ln 2}$$ At this point, we have the right-side function in the form of $e^{g(x)}$ where $g(x)=\sin x\ln 2$. This means we can use chain rule: $$\frac{dy}{dx}=\frac{d}{dg}e^{g(x)}\frac{d}{dx}\left(\sin x \ln 2\right)$$ It is a common identity that $\frac{d}{dg}e^{g(x)}=e^{g(x)}$. Also, it is a common identity that $\frac{d}{dx}\sin x=\cos x$, so we can use constant multiplication rule to get that $\frac{d}{dx}\left(\sin x \ln 2\right)=\ln 2\cdot \cos x$: $$\frac{dy}{dx}=e^{g(x)}\ln 2\cdot \cos x$$ Finally, substitute back in $g(x)=\sin x\ln 2$: $$\frac{dy}{dx}=e^{\sin x\ln 2}\ln 2\cdot \cos x$$ Now, to finish it off, we need to simplify. Note that $e^{\sin x \ln 2}=\left(e^{\ln 2}\right)^{\sin x}=2^{\sin x}$: $$\frac{dy}{dx}=2^{\sin x}\ln 2\cdot \cos x$$
• The way you have written it may appear that $\ln 2\cos x = \ln(2\cos x)$ Better to write it as $\ln 2\cdot \cos x$ or explicitly as $(\ln 2)(\cos x)$. Jun 30 '16 at 3:39
• @ParamanandSingh Thanks for the suggestion! I'm using $\ln 2\cdot\cos x$ now. Jun 30 '16 at 3:44
For this kind of problems, logarithmic differentiation makes life easier $$y=2^{\sin (x)}\implies \log(y)=\sin(x)\log(2)$$ Differentiating both sides $$\frac{y'}y=\cos(x)\log(2)\implies y'=2^{\sin (x)}\cos(x)\log(2)$$