Chain Rule — Question 9

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Question 9

Let f(x)=sin⁡2(5x3+1)f(x) = \sin^2(5x^3 + 1).

  • (a) Use the chain rule to compute f′(x)f'(x).

  • (b) Write down each layer of function composition clearly before differentiating.

Original worksheet page 1: question and worked solution for 3-9-009
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Question 9 - Solution

We are given: f(x)=sin⁡2(5x3+1)f(x) = \sin^2(5x^3 + 1)

This can be rewritten as: f(x)=[sin⁡(5x3+1)]2f(x) = [\sin(5x^3 + 1)]^2

Layers of function composition:

1. Innermost function: u(x)=5x3+1u(x) = 5x^3 + 1 2. Middle function: v(u)=sin⁡(u)v(u) = \sin(u) 3. Outer function: f(v)=v2f(v) = v^2

Differentiate step by step:

Start from the outside: f′(x)=2⋅sin⁡(5x3+1)⋅ddx[sin⁡(5x3+1)]f'(x) = 2 \cdot \sin(5x^3 + 1) \cdot \frac{d}{dx}[\sin(5x^3 + 1)]

Now differentiate the inner sine function: ddx[sin⁡(5x3+1)]=cos⁡(5x3+1)⋅ddx[5x3+1]=cos⁡(5x3+1)⋅15x2\frac{d}{dx}[\sin(5x^3 + 1)] = \cos(5x^3 + 1) \cdot \frac{d}{dx}[5x^3 + 1] = \cos(5x^3 + 1) \cdot 15x^2

Putting it all together: f′(x)=2⋅sin⁡(5x3+1)⋅cos⁡(5x3+1)⋅15x2f'(x) = 2 \cdot \sin(5x^3 + 1) \cdot \cos(5x^3 + 1) \cdot 15x^2

f′(x)=30x2⋅sin⁡(5x3+1)⋅cos⁡(5x3+1)\boxed{ f'(x) = 30x^2 \cdot \sin(5x^3 + 1) \cdot \cos(5x^3 + 1) }

Alternatively, you may express the result using the identity: 2sin⁡θcos⁡θ=sin⁡(2θ)2 \sin\theta \cos\theta = \sin(2\theta)

⇒f′(x)=15x2⋅sin⁡(2(5x3+1))=15x2⋅sin⁡(10x3+2)\Rightarrow f'(x) = 15x^2 \cdot \sin(2(5x^3 + 1)) = \boxed{15x^2 \cdot \sin(10x^3 + 2)}

Either boxed expression is valid.

Original worksheet page 2: question and worked solution for 3-9-009

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