Chain Rule — Question 3

Section 3-9 · Calculus pages 577–578

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Question 3

Let y=ln⁡(tan⁡3(5x2+1))y = \ln\left( \tan^3(5x^2 + 1) \right).

  • Differentiate yy with respect to xx using the chain rule.

  • Simplify the derivative as much as possible.

Question 3 - Solution

We are given: y=ln⁡(tan⁡3(5x2+1))y = \ln\left( \tan^3(5x^2 + 1) \right)

Step 1: Use logarithmic identity:

ln⁡(tan⁡3(5x2+1))=3ln⁡(tan(5x2+1))\ln\left( \tan^3(5x^2 + 1) \right) = 3 \ln\left( \tan(5x^2 + 1) \right)

Step 2: Differentiate using chain rule:

dydx=3⋅1tan⁡(5x2+1)⋅sec⁡2(5x2+1)⋅(10x)\frac{dy}{dx} = 3 \cdot \frac{1}{\tan(5x^2 + 1)} \cdot \sec^2(5x^2 + 1) \cdot (10x)

Step 3: Simplify the expression:

dydx=30xsec⁡2(5x2+1)tan⁡(5x2+1)\frac{dy}{dx} = \frac{30x \sec^2(5x^2 + 1)}{\tan(5x^2 + 1)}

dydx=30xsec⁡2(5x2+1)tan⁡(5x2+1)\boxed{ \frac{dy}{dx} = \frac{30x \sec^2(5x^2 + 1)}{\tan(5x^2 + 1)} }