Question

y=1/√5x find dy/dx full explanation of question​

Answer 1

Answer

• dy/dx = 1/2x√(5x)

Given

$\bullet \;\; \rm y=\dfrac{1}{\sqrt{5x}}$

To Find

$\bullet \;\; \rm \dfrac{dy}{dx}$

Solution

$\rm y=\dfrac{1}{\sqrt{5x}}\\\\\implies \rm y=\dfrac{1}{\sqrt{5}.\sqrt{x}}\\\\\implies \rm y=\dfrac{1}{\sqrt{5}}.\dfrac{1}{\sqrt{x}}$

On differentiating with respect to x on both sides , we get ,

$\implies \rm \dfrac{d}{dx}(y)=\dfrac{d}{dx}\bigg( \dfrac{1}{\sqrt{5}}.\dfrac{1}{\sqrt{x}}\bigg)\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{1}{\sqrt{5}}\dfrac{d}{dx}\bigg( \dfrac{1}{\sqrt{x}}\bigg)\\\\\rm Since,1/\sqrt{5}\ is\ constant.\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{1}{\sqrt{5}}\dfrac{d}{dx}\bigg( (\sqrt{x})^{-1}\bigg)\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{1}{\sqrt{5}}\dfrac{d}{dx}\bigg(x^{-\frac{1}{2}}\bigg)$

$\implies \rm \dfrac{dy}{dx}=\dfrac{1}{\sqrt{5}}\bigg( -\dfrac{1}{2}(x^{-\frac{1}{2}-1})\bigg)$

$\boxed{\rm Since,\dfrac{d}{dx}(x^n)=n.x^{n-1}}$

$\implies \rm \dfrac{dy}{dx}=\dfrac{1}{\sqrt{5}}\bigg( -\dfrac{1}{2}(x^{-\frac{3}{2}})\bigg)\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{1}{\sqrt{5}}\bigg( -\dfrac{1}{2x^{\frac{3}{2}}}\bigg)\\\\\implies \rm \dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{5}x^{\frac{3}{2}}}\\\\\implies \rm \dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{5}x.\sqrt{x}}}\\\\\implies \rm \dfrac{dy}{dx}=-\dfrac{1}{2x\sqrt{(5x)}}$