Question

plz do it fasttt...... ​

Answer 1

GIVEN :–

• A function $\bf { \tan}^{ - 1} (x)$.

TO FIND :–

• Differentiate form = ?

SOLUTION :–

• Let the function –

$\\ \implies\bf y = { \tan}^{ - 1} (x)$

• Let the function –

$\\ \implies\bf x= \tan( \theta)$

• Differentiate with respect to 'θ' –

$\\\implies\bf \dfrac{dx}{d \theta}= \sec^{2} ( \theta)$

$\\\implies\bf \dfrac{dx}{d \theta}= 1 + \tan^{2} ( \theta)$

$\\\implies\bf \dfrac{dx}{d \theta}= 1 +x^{2} \: \: \: \: \: - - - eq.(1)$

• Now –

$\\ \implies\bf y = { \tan}^{ - 1} ( \tan( \theta) )$

$\\ \implies\bf y = \theta$

• Differentiate with respect to 'θ' –

$\\ \implies\bf \dfrac{dy}{d \theta} = \dfrac{d\theta}{d\theta}$

$\\ \implies\bf \dfrac{dy}{d \theta} =1$

• We should write this as –

$\\ \implies\bf \dfrac{dy}{dx} \times \dfrac{dx}{d \theta} =1$

• Using eq.(1) –

$\\ \implies\bf \dfrac{dy}{dx} (1 + {x}^{2})=1$

$\\ \large \implies{ \boxed{\bf \dfrac{dy}{dx} = \dfrac{1}{(1+{x}^{2})}}}$

Answer 2

Answer:

SOLUTION :–

• Let the function –

$\\ \implies\bf y = { \tan}^{ - 1} (x)$

• Let the function –

$\\ \implies\bf x= \tan( \theta)$

• Differentiate with respect to 'θ' –

$\\\implies\bf \dfrac{dx}{d \theta}= \sec^{2} ( \theta)$

$\\\implies\bf \dfrac{dx}{d \theta}= 1 + \tan^{2} ( \theta)$

$\\\implies\bf \dfrac{dx}{d \theta}= 1 +x^{2} \: \: \: \: \: - - - eq.(1)$

• Now –

$\\ \implies\bf y = { \tan}^{ - 1} ( \tan( \theta) )$

$\\ \implies\bf y = \theta$

• Differentiate with respect to 'θ' –

$\\ \implies\bf \dfrac{dy}{d \theta} = \dfrac{d\theta}{d\theta}$

$\\ \implies\bf \dfrac{dy}{d \theta} =1$

• We should write this as –

$\\ \implies\bf \dfrac{dy}{dx} \times \dfrac{dx}{d \theta} =1$

• Using eq.(1) –

$\\ \implies\bf \dfrac{dy}{dx} (1 + {x}^{2})=1$

$\\ \large \implies{ \boxed{\bf \dfrac{dy}{dx} = \dfrac{1}{(1+{x}^{2})}}}$