If y-e4x sin 2x- then dy dx-

Here u=e^4x v=sin^2x, Differentiating both sides w.r.t. 'x' dy dx=d dx[e^4x sin2x] =e^4xd dx[sin 2x]+sin 2xd dx[e^4x](Using product rule) =2e^4xcos2x+sin2x 4e^ ...

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Byju's Answer

Standard XIII

Physics

Chain Rule of Differentiation

If y=e4x sin ...

Question

If $y = e^{4 x} s i n 2 x$ , then $\frac{d y}{d x}$ =?

$e^{4 x} c o s 2 x$

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$4 e^{4 x} s i n 2 x$

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$2 e^{4 x} [c o s 2 x + 2 s i n 2 x]$

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none of these

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Solution

The correct option is C
$2 e^{4 x} [c o s 2 x + 2 s i n 2 x]$

Here $u = e^{4 x} v = s i n^{2 x}$ , Differentiating both sides w.r.t. 'x'
$\frac{d y}{d x} = \frac{d}{d x} [e^{4 x} s i n 2 x]$
= $e^{4 x} \frac{d}{d x} [s i n 2 x] + s i n 2 x \frac{d}{d x} [e^{4 x}]$ (Using product rule)
= $2 e^{4 x} c o s 2 x + s i n 2 x 4 e^{4 x} = 2 e^{4 x} [c o s 2 x + 2 s i n 2 x]$

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If y=e4x sin2x, then dydx=?

The correct option is C 2e4x[cos2x+2sin2x] Here u=e4x v=sin2x, Differentiating both sides w.r.t. 'x' dydx=ddx[e4x sin2x] =e4xddx[sin 2x]+sin 2xddx[e4x](Using product rule) =2e4xcos2x+sin2x 4e4x=2e4x[cos2x+2sin2x]

If $y = e^{4 x} s i n 2 x$ , then $\frac{d y}{d x}$ =?