Question

differentiation of y= e to the power x (lnx.sinx+x square​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{dy}{dx}\:=\:e^x\:\left(\:\ln\:x\:sin\:x\:+\:\ln\:x\:\cos\:x\:+\:\dfrac{\sin\:x}{x}\:+\:x^2\:+\:2x\:\right)\:}}}$

Step-by-step-explanation:

The given function is

$\displaystyle{\sf\:y\:=\:e^x\:(\:ln\:x\:sin\:x\:+\:x^2\:)}$

We have to find the derivative of this function.

Now,

$\displaystyle{\sf\:y\:=\:e^x\:(\:\ln\:x\:sin\:x\:+\:x^2\:)}$

By using product rule of differentiation,

$\displaystyle{\sf\:(\:uv\:)'\:=\:u'\:v\:+\:u\:v'}$

Here,

$\displaystyle{\sf\:u\:=\:e^x}$

$\displaystyle{\sf\:v\:=\:\ln\:x\:\sin\:x\:+\:x^2}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{d}{dx}\:(\:e^x\:)\:(\:\ln\:x\:sin\:x\:+\:x^2\:)\:+\:\dfrac{d}{dx}\:(\:\ln\:x\:sin\:x\:+\:x^2\:)\:e^x}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:e^x\:(\:\ln\:x\:sin\:x\:+\:x^2\:)\:+\:e^x\:\left[\:\dfrac{d}{dx}\:(\:\ln\:x\:)\:\sin\:x\:+\:\dfrac{d}{dx}\:(\:\sin\:x\:)\:\ln\:x\:+\:\dfrac{d}{dx}\:(\:x^2\:)\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:e^x\:(\:\ln\:x\:sin\:x\:+\:x^2\:)\:+\:e^x\:\left(\:\dfrac{1}{x}\:\sin\:x\:+\:\cos\:x\:\ln\:x\:+\:2x\:\right)}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:e^x\:(\:\ln\:x\:sin\:x\:+\:x^2\:)\:+\:e^x\:\left(\:\dfrac{\sin\:x}{x}\:+\:\cos\:x\:\ln\:x\:+\:2x\:\right)}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:e^x\:\left(\:\ln\:x\:sin\:x\:+\:x^2\:+\:\dfrac{\sin\:x}{x}\:+\:\cos\:x\:\ln\:x\:+\:2x\:\right)}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:\dfrac{dy}{dx}\:=\:e^x\:\left(\:\ln\:x\:sin\:x\:+\:\ln\:x\:\cos\:x\:+\:\dfrac{\sin\:x}{x}\:+\:x^2\:+\:2x\:\right)\:}}}}$

Answer 2

$\longrightarrow$ In Given Question.

$y = e^x (In \: \: x \: \: sin \: \: x + x² )$

$\longrightarrow$ Now, Here we have to apply the Product rule Differentiation.

Hence,

$d \frac{uv}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

Where,

$\longrightarrow$ $\sf d \frac{uv}{dx}$ = Derivative Of $uv$ w.r.t. $x$

$\longrightarrow$ $\sf u$ = Variable $u$

$\longrightarrow$ $\sf \frac{dv}{dx}$ = Derivative of $v$ w.r.t. x

$\longrightarrow$ $\sf v$ = Variable $v$

$\longrightarrow$ $\sf \frac{du}{dx}$ = Derivative of $u$ w.r.t. x

$\longrightarrow$ $\sf du$ = Derivative $du$

Now,

$\longrightarrow$ According to question

• $\sf u = e^{x}$

• $\sf v = In \: \: x \: \: sin\: \: x + x²$

$\longrightarrow$ Now, we have to put the values

$\sf d \frac{uv}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$\sf \frac{dy}{dx} = (e^x) \frac{d}{dx} (In \: \: x \: \: sin \: \: x + x² ) + (In \: \: x \: \: sin \: \: x + x² ) \frac{d(e^x)}{dx}$

$\sf \frac{dy}{dx} = e^x (In \: \: x \: \: sin \: \: x + x²) + e^x \bigg[ \frac{d}{dx} (In \: \: x ) (sin\: \: x) + \frac{d}{dx} (sin\: \: x ) (In\: \: x) + \frac{d}{dx} (x²) \bigg]$

$\longrightarrow$ Differentiation of $\sf \frac{d}{dx} (sin \: \: x ) = cos \: \: x$

$\sf \frac{dy}{dx} = e^x (In \: \: x \: \: sin \: \: x + x² ) + e^x \bigg(\frac{1}{x} (sin\: \: x ) + cos \: \: x (In \: \: x ) + 2x \bigg)$

$\sf \frac{dy}{dx} = e^x (In \: \: x \: \: sin \: \: x + x² ) + e^x \bigg( \frac{sin\: \: x}{x} + cos \: \: x \: \: In \: \: x + 2x \bigg)$

$\longrightarrow$ Take common $e^x$ From the equation,

$\sf \underline{\boxed{ \frac{dy}{dx} = e^ x \bigg(In \: \: x \: \: sin \: \: x + x² + \frac{sin \: \: x}{x} + cos \: \: x \: \: In \: \: x + 2x \bigg) }}$

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I hope it helps you ❤️✔️