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.
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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.