Question

y-sin^-1(2x^3) =cos^-1(2x^3)​ find dy/dx

Answer 1

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

$\rm :\longmapsto\:\boxed{ \tt{ \: y - {sin}^{ - 1}( {2x}^{3}) = {cos}^{ - 1}( {2x}^{3})\: \: }}$

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

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

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

Given that,

$\rm :\longmapsto\:y - {sin}^{ - 1}( {2x}^{3}) = {cos}^{ - 1}( {2x}^{3})$

can be rewritten as

$\rm :\longmapsto\:y = {sin}^{ - 1}( {2x}^{3}) + {cos}^{ - 1}( {2x}^{3})$

We know,

$\boxed{ \tt{ \: {sin}^{ - 1}x + {cos}^{ - 1}x = \frac{\pi}{2} \: \: }}$

So, using this, we get

$\rm :\longmapsto\:y = \dfrac{\pi}{2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} \dfrac{\pi}{2}$

$\bf\implies \:\boxed{ \tt{ \: \dfrac{dy}{dx} = 0 \: }}$

Additional Information :-

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