Question

If y=
√1 + x²
then x? dy
is
dx​

Answer 1

Topic :-

Differentiation

Given :-

$\sf {y=\dfrac{x}{\sqrt{1+x^2}}}$

To Find :-

$\sf{x^3\cdot \dfrac{dy}{dx}}$

Solution :-

Calculating dy/dx,

$\sf {y=\dfrac{x}{\sqrt{1+x^2}}}$

$\sf {\dfrac{d(y)}{dx}=\dfrac{d}{dx}\left (\dfrac{x}{\sqrt{1+x^2}}\right )}$

$\sf{\dfrac{dy}{dx}=\dfrac{\sqrt{1+x^2}\cdot\dfrac{d(x)}{dx}-x\cdot \dfrac{d(\sqrt{1+x^2})}{dx}}{(\sqrt{1+x^2})^2}}$

$\sf {\left( \because \left( \dfrac{p}{q} \right)'=\dfrac{qp'-pq'}{q^2}\right)}$

$\sf{\dfrac{dy}{dx}=\dfrac{\sqrt{1+x^2}-x\cdot \dfrac{d(\sqrt{1+x^2})}{dx}}{(\sqrt{1+x^2})^2}}$

$\sf {\left ( \because \dfrac{dt}{dt}=1 \right )}$

$\sf{\dfrac{dy}{dx}=\dfrac{\sqrt{1+x^2}-x\cdot \dfrac{1}{2\sqrt{1+x^2}}\cdot\dfrac{d(1+x^2)}{dx}}{(\sqrt{1+x^2})^2}}$

$\sf {(\because (f(g))'=f'(g)\cdot g'\:\:and)}$

$\sf {\left (\because \dfrac{d(\sqrt{t})}{dt}=\dfrac{1}{2\sqrt{t}} \right)}$

$\sf{\dfrac{dy}{dx}=\dfrac{\sqrt{1+x^2}-x\cdot \dfrac{1}{2\sqrt{1+x^2}}\cdot2x}{(\sqrt{1+x^2})^2}}$

$\sf {\left(\because \dfrac{d(x^n)}{dx}=nx^{n-1}\right )}$

$\sf{\dfrac{dy}{dx}=\dfrac{\sqrt{1+x^2}-x\cdot \dfrac{1}{\not{2}\sqrt{1+x^2}}\cdot\not{2}x}{(\sqrt{1+x^2})^2}}$

$\sf{\dfrac{dy}{dx}=\dfrac{\sqrt{1+x^2}-\dfrac{x^2}{\sqrt{1+x^2}}}{(\sqrt{1+x^2})^2}}$

$\sf{\dfrac{dy}{dx}=\dfrac{\dfrac{(\sqrt{1+x^2})(\sqrt{1+x^2})-x^2}{\sqrt{1+x^2}}}{(\sqrt{1+x^2})^2}}$

$\sf{\dfrac{dy}{dx}=\dfrac{\dfrac{1+x^2-x^2}{\sqrt{1+x^2}}}{(\sqrt{1+x^2})^2}}$

$\sf {\dfrac{dy}{dx}=\dfrac{1}{(\sqrt{1+x^2})^2(\sqrt{1+x^2})}}$

$\sf {\dfrac{dy}{dx}=\dfrac{1}{(\sqrt{1+x^2})^3}}$

Calculating x³(dy/dx),

$\sf{x^3\cdot \dfrac{dy}{dx}}$

Substitute value of (dy/dx),

$\sf {x^3\cdot \dfrac{1}{(\sqrt{1+x^2})^3}}$

$\sf {\left ( \dfrac{x}{\sqrt{1+x^2}}\right )^3}$

$\sf{y^3}$

$\sf {\left (\because y=\dfrac{x}{\sqrt{1+x^2}}\right )}$

Answer :-

$\sf{x^3\cdot \dfrac{dy}{dx}=y^3}$