Question

Differentiate
$y = \sqrt{sin \: log \: e \: \sqrt{x} }$
by chain rule.​

Answer 1

Step-by-step explanation:

We have,

$\rm \: y = \sqrt{ \sin( log_{e}( \sqrt{x} ) ) }$

We know, $\rm\log_{e}(a)=ln(a)$, so,

$\rm \: y = \sqrt{ \sin( ln( \sqrt{x} ) ) }$

$\rm \: \implies \: \frac{dy}{dx} = \frac{1}{ 2\sqrt{ \sin( ln( \sqrt{x} ) ) } } . \frac{d}{dx} \{ \sin( ln( \sqrt{x} ) ) \}$

$\rm \: \implies \: \frac{dy}{dx} = \frac{1}{ 2\sqrt{ \sin( ln( \sqrt{x} ) ) } } . \cos( ln( \sqrt{x} ) ) . \frac{d}{dx} \{ ln( \sqrt{x} ) \}$

$\rm \: \implies \: \frac{dy}{dx} = \frac{1}{ 2\sqrt{ \sin( ln( \sqrt{x} ) ) } } . \cos( ln( \sqrt{x} ) ) . \frac{1}{\sqrt{x}} . \frac{d}{dx}( \sqrt{x} )$

$\rm \: \implies \: \frac{dy}{dx} = \frac{1}{ 2\sqrt{ \sin( ln( \sqrt{x} ) ) } } . \cos( ln( \sqrt{x} ) ) . \frac{1}{\sqrt{x}} . \frac{1}{2\sqrt{x} }$

$\rm \: \implies \: \frac{dy}{dx} = \frac{1}{ 4\sqrt{ \sin( ln( \sqrt{x} ) ) } } . \cos( ln( \sqrt{x} ) ) . \frac{1}{\sqrt{x}} . \frac{1}{\sqrt{x} }$

$\rm \: \implies \: \frac{dy}{dx} = \frac{1}{ 4\sqrt{ \sin( ln( \sqrt{x} ) ) } } . \cos( ln( \sqrt{x} ) ) . \frac{1}{x}$

$\rm \: \implies \: \frac{dy}{dx} = \frac{1}{ 4x\sqrt{ \sin( ln( \sqrt{x} ) ) } } . \cos( ln( \sqrt{x} ) )$

$\rm \: \implies \: \frac{dy}{dx} = \frac{\cos( ln( \sqrt{x} ) )}{ 4x\sqrt{ \sin( ln( \sqrt{x} ) ) } }$