Derivative of sin cube x | sin^3x Derivative

The derivative of sin cube x is equal to 3sin2x cosx. Here we learn how to differentiate sin^3x. The sin^3x derivative formula is d/dx(sin3x) = 3sin2x cosx.

The derivative of sine cube x is denoted by d/dx(sin3x) 0r (sin3x)$’$ and it is equal to

  • d/dx(sin3x) = 3sin2x cos x.
  • (sin3x)$’$ = 3sin2x cos x.

Derivative of sin3x by Chain Rule

To find the derivative of sine cube x by the chain rule of derivatives, let us put

z=sinx

∴ $\dfrac{dz}{dx}$ = cosx

Now, by the chain rule, the derivative of sin3x is equal to

$\dfrac{d}{dx}\big(\sin^3 x \big)$ = $\dfrac{d}{dz}(z^3)$ × $\dfrac{dz}{dx}$

= 3z2 × cosx by the power rule of derivatives and $\frac{dz}{dx}$ =cosx determined above.

= 3z2 cosx

= 3 sin2x cosx as z=sinx.

Therefore, the derivative of sin3x is equal to 3 sin2x cosx and this is obtained by the chain rule of derivatives.

ALSO READ:

Derivative of cos3xDerivative of cos2x

Derivative of sin3x by Product Rule

Note that sin3x can be written as a product of sinx and sin2x. So by the product rule, the derivative of sin cube x will be computed as follows:

$\dfrac{d}{dx}\big(\sin^3 x \big)$ = $\dfrac{d}{dx}\big(\sin x \cdot \sin^2 x \big)$

= $\sin x\dfrac{d}{dx}\big(\sin^2 x \big)$ + $\sin^2 x\dfrac{d}{dx}\big(\sin x \big)$

= sinx × 2sinx cosx + sin2x cosx

= 2sin2x cosx + sin2x cosx

= 3sin2x cosx.

So the differentiation of sin3x is equal to 3sin2x cosx and this is obtained by the product rule of differentiation.

FAQs

Q1: What is the derivative of sin^3x?

Answer: The derivative of sin3x is 3cosx sin2x.

Q2: Write the derivative formula of sin^3x.

Answer: The derivative formula of sin^3x is given by d/dx(sin3x) = 3cosx sin2x.

Q3: If y=sin^3x, then find dy/dx.

Answer: If y=sin3x, then dy/dx = 3sin2x cosx.

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