Question

if y=tan(sin^-1 x) then find dy/ dx​

Answer 1

❏Question :-

$\sf \: if \: y \: = \tan( { \sin}^{ - 1} x) \: \: then \: find \: \frac{dy}{dx}$

❏Answer :-

$\boxed{ \sf \frac{dy}{dx} = \frac{ { \sec}^{2} ( { \sin}^{ - 1} x)}{ \sqrt{1 - {x}^{2} } } } \:$

❏Used Formula :-

$\star \sf \: \frac{d}{dx} \tan x = { \sec}^{2} x \\ \\ \star \sf \: \frac{d}{dx} \: { \sin}^{ - 1} x = \frac{1}{ \sqrt{1 - {x}^{2} } }$

❏Solution :-

We have ,

$\sf \mapsto \: y \: = \tan( { \sin}^{ - 1} x) \: \: \\ \\ \bf \red{ differentiate \: with \:} respect \: to \: x \\ \\ \mapsto \sf \frac{dy}{dx} = \frac{d}{dx}\tan( { \sin}^{ - 1} x) \: \\ \\ \mapsto \: \sf \frac{dy}{dx} = { \sec}^{2} ( { \sin}^{ - 1} x) \: \frac{d}{dx} ( { \sin}^{ - 1} x) \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\because \sf \frac{d}{dx} ( { \sin}^{ - 1} x) = \frac{1}{ \sqrt{1 - {x}^{2} } } } \\ \therefore \: \\ \: \\ \mapsto \: \sf \frac{dy}{dx} = { \sec}^{2} ( { \sin}^{ - 1} x) \: \frac{1}{ \sqrt{1 - {x}^{2} } } \\ \\ \mapsto \boxed{ \sf \frac{dy}{dx} = \frac{ { \sec}^{2} ( { \sin}^{ - 1} x)}{ \sqrt{1 - {x}^{2} } } }$

$\boxed{\begin{minipage}{7 cm}\rm \red{Some \: same \: questions \: for \: practice }\\ \\</p><p>(1)\rm If \: y = \sin(\tan^{-1}x) \:,find \:\frac{dy}{dx} \\ \\ </p><p>(2)\rm If \: y = 2x^{2} + 3x + 5 \: ,find \: \frac{dy}{dx} \\ \\ </p><p>(3)\rm if , \: y = x^{x} \: , \:find \: \frac{dy}{dx}\\ \\ </p><p>\end{minipage}}$