Question

If given that y = x ( arcsin x ) + √( 1 - x² ) . Prove thT y' = arcsin x​

Answer 1

Topic - Differentiation

Explanation:

We are given that $\rm y = x(sin^{-1}x) + \sqrt{1-x^2}$. We need to prove that $\rm y' = sin^{-1}x$.

We will use the following rules of derivative to solve the given problem:

$\boxed{\begin{array}{lcc}\rm\dfrac{d}{dx}\{x^n\} = n\cdot x^{n-1} \\ \\\rm\dfrac{d}{dx} \{\sqrt{x}\} = \dfrac{1}{2\sqrt{x}}\\\\ \rm \dfrac{d}{dx} \{f(x) \pm g(x) \} = \dfrac{d}{dx} \{ f(x)\} \pm \dfrac{d}{dx} \{ g(x)\} \\ \\ \rm \dfrac{d}{dx} \{f(x) \cdot g(x)\} = f'(x) \cdot g(x) + f(x) \cdot g'(x) \\\\ \rm\dfrac{d}{dx} \{f(g(x))\}= f'(g(x)) \cdot g'(x)\\\\ \rm\dfrac{d}{dx} \{sin^{-1}x\}=\dfrac{1}{\sqrt{1-x^2}}\end{array}}$

So let's solve the given problem!

We have,

$\tt\implies y \prime = x( {sin}^{ - 1} (x)) + \sqrt{1 - {x}^{2} }$

Differentiation both sides w.r.t. x will give us,

$\tt \implies y \prime= \dfrac{d}{dx} \left \{x( {sin}^{ - 1} (x)) + \sqrt{1 - {x}^{2} } \right \}$

$\tt \implies y \prime= \dfrac{d}{dx} \left \{x( {sin}^{ - 1} (x)) \right\} + \dfrac{d}{dx} \left \{ \sqrt{1 - {x}^{2} } \right \}$

$\tt \implies y \prime= {sin}^{ - 1} (x) \cdot \dfrac{d}{dx} \{x \} + x \cdot \dfrac{d}{dx} \{ {sin}^{ - 1}(x) \} + \dfrac{1}{2 \sqrt{1 - {x}^{2} }} \cdot \dfrac{d}{dx} \{ 1 - {x}^{2} \}$

$\tt \implies y \prime= {sin}^{ - 1} (x) \cdot (1) + x \cdot \dfrac{1}{ \sqrt{1 - {x}^{2} } } + \dfrac{1}{2 \sqrt{1 - {x}^{2} }} \cdot \left(\dfrac{d}{dx} \{ 1 \} - \dfrac{d}{dx} \{ {x}^{2} \} \right)$

$\tt \implies y \prime= {sin}^{ - 1} (x) + \dfrac{x}{ \sqrt{1 - {x}^{2} } } + \dfrac{1}{2 \sqrt{1 - {x}^{2} }} \left( 0 - 2x\right)$

$\tt \implies y \prime= {sin}^{ - 1} (x) + \dfrac{x}{ \sqrt{1 - {x}^{2} } } + \dfrac{1}{ \not2 \sqrt{1 - {x}^{2} }} \left( - \not2x\right)$

$\tt \implies y \prime= {sin}^{ - 1} (x) + \dfrac{x}{ \sqrt{1 - {x}^{2} } } - \dfrac{ x}{ \sqrt{1 - {x}^{2} }}$

$\boxed{\red{ \tt\implies y \prime= {sin}^{ - 1} (x)}}$

Hence proved.