Question

Derivative of 1 upon 1+coax.dx=?

Answer 1

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

Given function is

$\rm :\longmapsto\: \dfrac{1}{1 + cosax}$

Let assume that

$\rm :\longmapsto\:y \: = \: \dfrac{1}{1 + cosax}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y \: = \dfrac{d}{dx}\: \dfrac{1}{1 + cosax}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{d}{dx} {(1 + cosax)}^{ - 1}$

We know,

$\purple{\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - 1 \: {(1 + cosax)}^{ - 1 - 1}\dfrac{d}{dx}(1 + cosax)$

We know,

$\blue{\boxed{ \rm{ \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v}}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: {(1 + cosax)}^{ -2} \bigg(\dfrac{d}{dx}1 + \dfrac{d}{dx}cosax \bigg)$

We know,

$\green{\boxed{ \rm{ \dfrac{d}{dx}k = 0}}}$

and

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

So, using these, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: {(1 + cosax)}^{ -2} \bigg(0 - sinax\dfrac{d}{dx}ax \bigg)$

We know,

$\red{\boxed{ \rm{ \dfrac{d}{dx}k \: f(x) = \: k \: \dfrac{d}{dx}f(x)}}}$

So, using this

$\rm :\longmapsto\:\dfrac{dy}{dx} =sinax \: {(1 + cosax)}^{ -2} \bigg( a \: \dfrac{d}{dx}x \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =a \: sinax \: {(1 + cosax)}^{ -2} \bigg( \: 1 \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =a \: sinax \: {(1 + cosax)}^{ -2}$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{a \: sinax}{ {(1 + cosax)}^{2} }$

Additional Information :-

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

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

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

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

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

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

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