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

Find the derivative of the following functions
$y={2}^{3+\mathrm{sin}x}$
You can still ask an expert for help

## Want to know more about Derivatives?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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

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

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

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

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

Jeffrey Jordon