Question

differentiate with respect to x​

Answer 1

Answer:

1/2.1/√sec√xsec√xtan√x.1/2√x

Answer 2

$\large\underline{\sf{Solution-}}$

To differentiate,

$\red{\rm :\longmapsto\: \sqrt{sec \sqrt{x} } }$

Let assume that

$\rm :\longmapsto\:y = \sqrt{sec \sqrt{x} }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} \sqrt{sec \sqrt{x} }$

We know,

$\boxed{ \rm{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{1}{2 \sqrt{sec \sqrt{x} } } \dfrac{d}{dx} {sec \sqrt{x} }$

We know,

$\boxed{ \rm{ \dfrac{d}{dx}secx = secx \: tanx}}$

So, using this,

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{1}{2 \sqrt{sec \sqrt{x} } } {sec \sqrt{x} } \: tan \sqrt{x} \dfrac{d}{dx} \sqrt{x}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{1}{2 \sqrt{sec \sqrt{x} } } {( \sqrt{sec \sqrt{x} } ) \: }^{2} \: tan \sqrt{x} \dfrac{d}{dx} \sqrt{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{2} \sqrt{sec \sqrt{x} } \: tan \sqrt{x} \times \dfrac{1}{2 \sqrt{x} }$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{tan \sqrt{x} \: \sqrt{sec \sqrt{x} } }{4 \sqrt{x} }$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosx = - sinx}}$

$\boxed{ \rm{ \dfrac{d}{dx}tanx = {sec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cotx = { - cosec}^{2}x}}$

$\boxed{ \rm{ \dfrac{d}{dx}cosecx \: = \: - \: cosecx \: cotx}}$

$\boxed{ \rm{ \dfrac{d}{dx}logx = \frac{1}{x}}}$