Question

differentiate y=cos^-1(4cos^3x-3cosx) wrt x

Answer 1

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

$\rm :\longmapsto\:y = {cos}^{ - 1}[ 4{cos}^{3}x - 3cosx]$

$\red{\large\underline{\sf{To\:Find - }}}$

$\rm :\longmapsto\:\dfrac{dy}{dx}$

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

Given function is

$\rm :\longmapsto\:y = {cos}^{ - 1}[ 4{cos}^{3}x - 3cosx]$

We know that,

$\purple{\boxed{ \tt{ \: {4cos}^{3}x - 3cosx = cos3x \: }}}$

So, above function can be rewritten as

$\rm :\longmapsto\:y = {cos}^{ - 1}[cosx]$

$\rm :\longmapsto\:y = 3x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}3x$

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} =3 \dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} =3 \times 1$

$\purple{\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} = 3 \: }}}$

More to know :-

$\green{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$