Differentiate
$f (x) = ln (1 + x^{2})$

Open in App

Solution

Given $f (x) = l n (1 + x^{2})$
Differentiate on both sides w.r.t x
$\frac{d f (x)}{d x} = \frac{1}{1 + x^{2}} \frac{d}{d x} (1 + x^{2})$
$f^{'} (x) = \frac{2 x}{1 + x^{2}}$
$∴ f^{'} (x) = \frac{2 x}{1 + x^{2}}$

Suggest Corrections

Similar questions

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

f (\frac{2 x}{1 + x^{2}})

f (x) = \sqrt{\frac{1 + x^{2}}{1 - x^{2}}}

f (x) = {log}_{e} \sqrt{4 - x^{2}}

\int \frac{(2 x + 3) (x^{2} - 3 x + 1) d x}{(x^{4} - 7 x^{2} + 1) (1 + ln (1 + 3 x + x^{2}))} = f (x) + C,

C

f (0) = 0,

f (x) = \int_{0}^{x} l n (\frac{1 - t}{1 + t}) d t \Rightarrow

Join BYJU'S Learning Program