CBSE Class 12 · DifferentiationLong Answer

f (x) = \frac{sec x - 1}{sec x + 1}

, find

f (x)

in simplified form, and also find

f^{'} (\frac{π}{2})

Answer: $f (x) = tan \frac{x}{2}, f^{'} (π /2) = 1$

Solution

Multiply inside by

cos x

\frac{sec x - 1}{sec x + 1} = \frac{1 - cos x}{1 + cos x} = tan^{2} \frac{x}{2}

.
Hence, for

x

where

tan \frac{x}{2} > 0

f (x) = tan \frac{x}{2}

Differentiate by the chain rule:

f^{'} (x) = \frac{1}{2} sec^{2} \frac{x}{2}

x = \frac{π}{2}

\frac{x}{2} = \frac{π}{4}

and

sec^{2} \frac{π}{4} = 2

f^{'} (\frac{π}{2}) = \frac{1}{2} \cdot 2 = 1

f (x) = tan \frac{x}{2}, f^{'} (\frac{π}{2}) = 1

CBSE Class 12 Board · Subjective