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How do you find the derivative of the function $si n^{3} (2 x)$ ?

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6 si n^{2} (2 x) \cos (2 x)

6 si n^{2} (2 x) co s^{2} (2 x)

6 si n^{3} (2 x) \cos (2 x)

6 si n^{2} (2 x) \cos (x)

answer is A.

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Detailed Solution

Let’s assume

\sin 2 x = z

and

2 x = y

.
So,

p = si n^{3} (2 x) \Leftrightarrow p = z^{3}

……. 1
So, let us differentiate both sides of equation (1) w.r.t

x .

From the chain rule we get,

\frac{d (p)}{dx} = \frac{d (z^{3})}{dx}

\Rightarrow

\frac{dp}{dx} = \frac{d {(z}^{3})}{dz} \times \frac{dz}{dx}

\Rightarrow

\frac{dp}{dx} = 3 z^{2} \times \frac{dz}{dx}

[

\frac{d (x^{n)}}{dx} = n x^{n - 1}

]

\Rightarrow

\frac{dp}{dx} = 3 si n^{2} (2 x) \times \frac{d (sin y)}{dx}

……… 2 [Since,

\sin 2 x = z

and

2 x = y

]
Let's Apply chain rule in equation (2), we get

\Rightarrow

\frac{dp}{dx} = 3 si n^{2} (2 x) \times \frac{d (sin y)}{dy} \times \frac{dy}{dx}

\Rightarrow

\frac{dp}{dx} = 3 si n^{2} (2 x) \times cos y \times \frac{dy}{dx}

[

\frac{d (sin x)}{dx} = cos x]

Let’s put

2 x = y

\Rightarrow

\frac{dp}{dx} = 3 si n^{2} (2 x) \times \cos (2 x) \times \frac{d (2 x)}{dx}

\Rightarrow

\frac{dp}{dx} = 3 si n^{2} (2 x) \times \cos (2 x) \times 2

[

\frac{d (ax)}{dx} = a]

\Rightarrow \frac{dp}{dx} = 6 si n^{2} (2 x) \cos (2 x)

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