Get the answer to "Find the derivative of the following functions y=23+sin⁡x"

Marvin Mccormick

Answered question

2021-10-08

Find the derivative of the following functions

y = 2^{3 + \sin x}

Answer & Explanation

Adnaan Franks

Skilled2021-10-09Added 92 answers

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

$\frac{1}{y} \frac{d y}{d x} = \ln (2) \frac{d}{d x} [(3 + \sin x)]$

$\frac{1}{y} \frac{d y}{d x} = \ln (2) \cdot (0 + \cos x)$

$\frac{1}{y} \frac{d y}{d x} = \ln (2) \cdot \cos x$

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

Jeffrey Jordon

Expert2022-03-24Added 2605 answers

Answer is given below (on video)

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