Find the derivative of the following functions y=2^{3+\sin x}

Step 1
Consider the provided function,
${\displaystyle y=2^{3+\sin x}}$...(1)
Find the derivative of the above functions.
First, we taking log both the sides.
${\displaystyle \ln y=\ln \left(2^{3+\sin x}\right)}$
Apply the log property ${\displaystyle \ln \left(m^n\right)=n\ln \left(m\right)}$.
So,
${\displaystyle \ln y=\ln \left(2^{3+\sin x}\right)}$
${\displaystyle \ln y=(3+\sin x)\ln \left(2\right)}$
Step 2
Now, the above equation differentiate with respect to x.
Apply the common derivative rule ${\displaystyle \frac{d}{dx}\left(\ln x\right)=\frac{1}{x}\cdot 1\text{and}\frac{d}{dx}\left(\sin x\right)=\cos x}$
${\displaystyle \frac{d}{dx}\left(\ln y\right)=\frac{d}{dx}\left[(3+\sin x)\ln \left(2\right)\right]}$

${\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \left(2\right)\frac{d}{dx}\left[(3+\sin x)\right]}$

${\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \left(2\right)\cdot (0+\cos x)}$

${\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \left(2\right)\cdot \cos x}$

${\displaystyle \frac{dy}{dx}=y\cdot \ln \left(2\right)\cdot \cos x}$
Step 3
Now, in the equation (1) we put the value of y in the above derivative.
We get,
${\displaystyle \frac{dy}{dx}=\left(2^{3+\sin x}\right)\cdot \ln \left(2\right)\cdot \cos x}$
${\displaystyle =\ln \left(2\right)\left(2^{3+\sin x}\right)\cos x}$
Hence.