Question

y= Sin 3x.Sin 8x find dy/dx​

Answer 1

ANSWER :–

$\\ \dashrightarrow { \bf \dfrac{dy}{dx} =8 \sin(3x)\cos(8x) + 3 \sin(8x)( \cos 3x)}$

EXPLANATION :–

GIVEN :–

$\\ \bf \implies y = \sin(3x) . \sin(8x)$

TO FIND :–

$\\ \bf \implies \dfrac{dy}{dx} =?$

SOLUTION :–

$\\ \bf \implies y = \sin(3x) . \sin(8x)$

• We know that –

$\\ \bf \implies \dfrac{d(u.v)}{dx} =u \dfrac{dv}{dx} + v \dfrac{du}{dx}$

• And –

$\\ \bf \implies \dfrac{d( \sin x)}{dx} = \cos x$

• So that –

$\\ \bf \implies \dfrac{dy}{dx} = \sin(3x) \dfrac{d( \sin 8x) }{dx} + \sin(8x) \dfrac{d( \sin 3x) }{dx}$

$\\ \bf \implies \dfrac{dy}{dx} = \sin(3x)(8) \cos(8x) + \sin(8x) (3)( \cos 3x)$

$\\ \implies { \boxed{ \bf \dfrac{dy}{dx} =8 \sin(3x)\cos(8x) + 3 \sin(8x)( \cos 3x)}}$

$\\ \rule{220}{2}$

Answer 2

Given ,

The function is

• Y = Sin(3x) . Sin(8x)

Differentiating with respect to x , we get

$\sf \mapsto \frac{dy}{dx} = \frac{d}{dx} Sin(3x) . Sin(8x) \\ \\ \sf \mapsto \frac{dy}{dx} = Sin(8x) \frac{d \{Sin(3x) \}}{dx} + Sin(3x) \frac{d \{Sin(8x) \}}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} = Sin(8x) \times 3 \{Cos(3x) \} + Sin(3x) \times 8 \{ Cos(8x) \} \\ \\ \sf \mapsto \frac{dy}{dx} = 8\{ Cos(8x) \} \times Sin(3x) + 3\{Cos(3x) \} \times Sin(8x)$

Remmember :

$\sf \mapsto \frac{d(u.v)}{dx} = v \frac{d(u)}{dx} + u \frac{(v)}{dx} </p><p>$