Question

y=x/(x^2+r^2)^3/2 if dy/dx=0 then find x

Answer 1

ANSWER :–

$\\ \: { \: { \: { x = \frac{ r }{ \sqrt{2} } }}}$

EXPLANATION :–

GIVEN :–

• A function – ${ \boxed{ \bold { \: \: y = \frac{x}{ {( {x}^{2} + {r}^{2}) }^{ \frac{3}{2} } }}}}$

• And ${ \bold{ \frac{dy}{dx} = 0 }}$

TO FIND :–

• Value of 'x'

SOLUTION :–

• According to the question –

${ \bold{ \: \: \: \:\implies y = \frac{x}{ {( {x}^{2} + {r}^{2}) }^{ \frac{3}{2} } }}}$

• Now Differentiate with respect to 'x' –

$\\ { \bold{ \implies \frac{dy}{dx} = \frac{ {( {x}^{2} + {r}^{2} )}^{ \frac{3}{2} } (1) - x[ \frac{3}{2} {( {x}^{2} + {r}^{2} ) }^{ \frac{1}{2} }](2x) }{ {( {x}^{2} + {r}^{2}) }^{ \frac{3}{2} } }}}$

$\\ \: { \bold{ \implies \frac{dy}{dx} = \frac{ {( {x}^{2} + {r}^{2} )}^{ \frac{1}{2} } [( {x}^{2} + {r}^{2}) - 3 {x}^{2}] }{ {( {x}^{2} + {r}^{2}) } }}}$

$\\ \: { \bold{ \implies \frac{dy}{dx} = \frac{ {( {x}^{2} + {r}^{2} )}^{ \frac{1}{2} } [ {r}^{2} - 2 {x}^{2}] }{ {( {x}^{2} + {r}^{2}) } }}}$

$\\ \: { \bold{ \implies \frac{dy}{dx} = \frac{ {r}^{2} - 2 {x}^{2} }{ {( {x}^{2} + {r}^{2})^{ \frac{5}{2} } } }}}$

• But According to question –

${ \bold{ \implies \frac{dy}{dx} = 0 }}$

$\\ \: { \bold{ \implies \frac{ {r}^{2} - 2 {x}^{2} }{ {( {x}^{2} + {r}^{2})^{ \frac{5}{2} } } } = 0}}$

$\\ \: { \bold{ \implies {r}^{2} - 2 {x}^{2} = 0 }}$

$\\ \: { \bold{ \implies {r}^{2} = 2 {x}^{2} }}$

$\\ \: { \implies \: { \boxed{ \: \bold{ x = \frac{ r }{ \sqrt{2} } }}}}$

USED FORMULA :–

$\\ { \bold{ (1) \: \: \frac{d( {x}^{n} )}{dx} = n {x}^{n - 1} }}$

$\\ { \bold{ (2) \: \: \frac{d}{dx}( \frac{u}{v}) = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {v}^{2} } }}$