Question

differentiate log(sec x/2 + tan x/2)

Answer 1

y = log(secx/2 + tanx/2)

Using chain rule
let secx/2+tanx/2 = t
Now dt/dx = 1/2(secx/2Tanx/2 + sec²x/2)
Now y = log t
dy/dt = 1/t
Now dy/dx = dy/dt × dt/dx
= 1/secx/2+tanx/2 × 1/2(secx/2Tanx/2+sec²x/2)

If you will face any problem then comments below .

Answer 2

$\left(\log \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)\right=\sec \frac{\mathrm{x}}{2}$

In maths the “differential calculus” is a “subfield” of the calculus concerned with study of the rate at which the quantities change. It is one of the traditional ways of calculus

Given:

$\left(\log \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)\right$

To find:

Differentiate the given value

Answer:

Given equation is,

$\left(\log \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)\right.$

$=\frac{d}{d x}\left(\log \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)\right.$

Using chain rule,

$f(x)=\frac{1}{f(x)} f^{\prime}(x)$

We get,

$\frac{d}{d x}\left(\log \left(\sec \frac{\mathrm{x}}{2}+\tan \frac{\mathrm{x}}{2}\right)=\frac{1}{\left(\sec \frac{\mathrm{x}}{2}+\tan \frac{\mathrm{x}}{2}\right)}\left(\sec \frac{\mathrm{x}}{2}+\tan \frac{\mathrm{x}}{2}\right)\right.$

$=\frac{1}{\left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)}\left(\sec \frac{x}{2} \tan \frac{x}{2}+\sec ^{2}\left(\frac{x}{2}\right)\right)$

$=\frac{1}{\left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)}\left(\sec \frac{x}{2}\left(\tan \frac{x}{2}+\sec \left(\frac{x}{2}\right)\right)\right)$

$\left(\log \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)\right=\sec \frac{\mathrm{x}}{2}$