Question

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$\sf★ Find \: \frac{dy}{dx} \\ \sf \: if \: y = {sin \: x}^{(tan \: x)}$

◇ Give stepwise answer​

Answer 1

Answer:

Given:-

• $y = {sin \: x}^{(tan \: x)}$

To Find:-

• Value of $\dfrac{dy}{dx}$

Solution:-

As given,

⟾ $y = {sin \: x}^{(tan \: x)}$

On taking log both side,

⟾ log y = tan x log sin x

Diff.w.r.to.x

⟿ $\dfrac{1}{y}×\dfrac{dy}{dx}\: = \:tanx(\dfrac{d}{dx\:log sin\:x}) \:+\:log sin x(\dfrac{d}{dx\:tan\:x}$)

⟿ $\dfrac{1}{y}×\dfrac{dy}{dx}\: =\: tanx \dfrac{1}{sinx}.\dfrac{d}{dxsin\: x}+log \:sin\:x sec^{2x}$

⟿ $\dfrac{1}{y}×\dfrac{dy}{dx}\:= \:tanx \dfrac{1}{sin\:x.cos\:x} + log\:sin\:x sec^{2x}$

⟿ $\dfrac{1}{y}×\dfrac{dy}{dx}$ =$\dfrac{tanx.cosx}{sinx}+log\:sin\:x sec^{2x}$

⟿ $\dfrac{1}{y}×\dfrac{dy}{dx}$ =$\dfrac{sinx}{cosx}.\dfrac{cosx}{sin\:x} +log\:sin\:x sec^{2x}$

⟿ $\dfrac{1}{y}×\dfrac{dy}{dx}$ = $1+log sinx sec^{2x}$

⟿ $\dfrac{dy}{dx}$ = $y(1+log \:sin\:x sec^{2x})$

⟿ $\dfrac{dy}{dx}$ = $sinx^{(tan\:x)}[1+log\:sin\:x sec^{2x}]$

Hence, the Value of ${\bold{ \dfrac{dy}{dx}}}$ is

${\bold{sinx^{(tan\:x)}[1+log\:sin\:x sec^{2x}]}}$.

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Answer 2

Answer:

Step-by-step explanation:

$</p><p> \red{ \begin{gathered}\sf {x}^{y}{y}^{x} = {x+y}^{x+y} \\ \\ \sf Taking\:log\:on\:both\:sides \\ \\ \sf log {x}^{y}{y}^{x} = log {x+y}^{x+y} \\ \\ \sf log{x}^{y} + log{y}^{x} = (x+y)log(x+y) \\ \\ \sf ylogx + xlogy = (x+y)log(x+y) \\ \\ \sf Now\: differentiating\:w.r.t\: x \\ \\ \sf \frac{dy}{dx}logx + \frac{y}{x} + logy + \frac{x}{y} \frac{dy}{dx} = (1+\frac{dy}{dx})log(x+y) + \frac{x+y}{x+y}(1+\frac{dy}{dx}) \\ \\ \sf \frac{dy}{dx}logx + \frac{y}{x} + logy + \frac{x}{y} \frac{dy}{dx} = log(x+y) + \frac{dy}{dx}log(x+y) + 1+\frac{dy}{dx} \\ \\ \sf \frac{dy}{dx}logx+ \frac{x}{y} \frac{dy}{dx} - \frac{dy}{dx}log(x+y) - \frac{dy}{dx} = log(x+y) + 1 - logy - \frac{y}{x} \\ \\ \sf \frac{dy}{dx}(logx + \frac{x}{y} - log(x+y) - 1 ) = log(x+y) + 1 - logy - \frac{y}{x} \\ \\ \sf \frac{dy}{dx} = \frac{log(x+y) + 1 - logy - \frac{y}{x}}{logx + \frac{x}{y} - log(x+y) - 1 } \end{gathered}}</p><p>$