"Find the derivatives of the functions f(0)=(sin⁡01+cos⁡0)2" was answered by students like you

ka1leE

Answered question

2021-09-22

Find the derivatives of the functions

f (0) = {(\frac{\sin 0}{1 + \cos 0})}^{2}

pattererX

Skilled2021-09-23Added 95 answers

Given information:
The function is

f (0) = {(\frac{\sin 0}{1 + \cos 0})}^{2}

.
Calculate the derivative of the function as shown below.

\frac{d f}{d 0} = 2 {(\frac{\sin 0}{1 + \cos 0})}^{2 - 1} \times \frac{d}{d x} (\frac{\sin 0}{1 + \cos 0})

= 2 (\frac{\sin 0}{1 + \cos 0}) \frac{(1 + \cos 0) \times \frac{d}{d x} \sin 0 - \sin 0 \times \frac{d}{d x} (1 + \cos 0)}{{[1 + \cos 0]}^{2}}

= 2 (\frac{\sin 0}{1 + \cos 0}) \frac{(1 + \cos 0) \times \cos 0 - \sin 0 \times (0 - \sin 0)}{{[1 + \cos 0]}^{2}}

= \frac{2 \sin 0 (\cos 0 + \cos^{2} 0 + \sin^{2} 0)}{{[1 + \cos 0]}^{3}}

= \frac{2 \sin 0 (\cos 0 + 1)}{{[1 + \cos 0]}^{3}}

= \frac{2 \sin 0}{{(1 + \cos 0)}^{2}}

Therefore, the derivative of the function

(\frac{d f}{d 0}) i s \frac{2 \sin 0}{{(1 + \cos 0)}^{2}}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?