Question

Find the derivative of f(x)=cotx ln secx

Answer 1

Step-by-step explanation:

$f(x) = \cot(x) ln( \sec(x) \\ \\ {f(x)}^{1} = \frac{ \cot(x) }{ \sec(x) } \sec(x) \tan(x) - { \csc(x) }^{2} ln( \sec(x)) \\ \\ = 1 - { \csc(x) }^{2} ln( \sec(x) )$

Answer 2

Topic :-

Differentiation

To Differentiate :-

$f(x)=\cot x \cdot \ln \sec x$

Solution :-

$f(x)=\cot x \cdot \ln \sec x$

$\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)$

$\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}$

$\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}$

$\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x$

$\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)$

$\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1$

$(\because \cot x \cdot \tan x = 1)$

Answer :-

$\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}$

Note : csc x = cosec x