Question

if y=sin(4x + 2Π/3) find the value of dy/dx

Answer 1

dy/dx = [cos {4x+(2π/3)}] × 4
dy/dx = 4 × [cos {(12x + 2π)/3}]

Answer 2

Answer:

Required value of $\frac{dy}{dx}$ is $4 \cos(4x + \frac{2\pi}{3})$

Explanation:

Given,

$y = \sin(4x + \frac{2\pi}{3} )$

Here we want to find value of $\frac{dy}{dx}$

We are differentiating both side with respect to x,

$\frac{dy}{dx} = \frac{d}{dx} ( \sin(4x + \frac{2\pi}{3} ) ) \\ = \cos(4x + \frac{2\pi}{3}) \times \frac{d}{dx} (4x + \frac{2\pi}{3}) \\ = \frac{d}{dx} (4x) + \frac{d}{dx} ( \frac{2\pi}{3} ) \cos(4x + \frac{2\pi}{3}) \\ =( 4 + 0) \cos(4x + \frac{2\pi}{3}) \\ = 4 \cos(4x + \frac{2\pi}{3})$

Here applied formulas are

$\frac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1} \\ \frac{d}{dx} (sinx) = cosx \\ \frac{d}{dx} (constant) = 0$

Some important Derivatives formula

$1. \frac{d}{dx} ( \sin(x) ) = \cos(x ) \\ 2. \frac{d}{dx} ( \cos(x) ) = - \sin(x) \\ 3. \frac{d}{dx} ( \tan(x) ) = {sec}^{2} x \\ 4. \frac{d}{dx} ( \cot(x) ) = - {cosec}^{2} x \\ 5. \frac{d}{dx} ( \sec(x) ) = \sec(x) \tan(x) \\ 6. \frac{d}{dx} (cosec(x)) = - cosecxcotx$

Know more about Derivative

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