If y-e2x-cosln x- then dydx is

Given, nbsp;y=e^2^x cos(ln x) So, we have using chain rule, dydx=d(e^2^x cos(ln x))dx ⇒ de^2^xd2^x×d2^xdx dcos(ln x)dln x×dln xdx ⇒ 2^xe^2^xln 2+sin(ln x)x ...

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Byju's Answer

Standard XIII

Physics

Basic Differentiation Rule

If y=e2x-cosl...

Question

If $y = e^{2^{x}} - cos (ln x)$ , then $\frac{d y}{d x}$ is

e^{2^{x}} + \frac{sin (ln x)}{x}

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2^{x} e^{2^{x}} ln 2 + \frac{sin (ln x)}{x}

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e^{2^{x}} + sin (ln x)

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2^{x} e^{2^{x}} ln 2 - sin (ln x)

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Solution

The correct option is B $2^{x} e^{2^{x}} ln 2 + \frac{sin (ln x)}{x}$
Given, $y = e^{2^{x}} - cos (ln x)$
So, we have using chain rule,
$\frac{d y}{d x} = \frac{d (e^{2^{x}} - cos (ln x))}{d x}$
$\Rightarrow \frac{d e^{2^{x}}}{d 2^{x}} \times \frac{d 2^{x}}{d x} - \frac{d cos (ln x)}{d ln x} \times \frac{d ln x}{d x}$
$\Rightarrow 2^{x} e^{2^{x}} ln 2 + \frac{sin (ln x)}{x}$

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If y=e2x−cos(lnx), then dydx is

The correct option is B 2xe2xln2+sin(lnx)xGiven, y=e2x−cos(lnx) So, we have using chain rule, dydx=d(e2x−cos(lnx))dx ⇒ de2xd2x×d2xdx−dcos(lnx)dlnx×dlnxdx ⇒ 2xe2xln2+sin(lnx)x

If $y = e^{2^{x}} - cos (ln x)$ , then $\frac{d y}{d x}$ is