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Exercise 10.3 · Q16

Q.Differentiate the following: s(t)=t3+1t3−14s(t) = \sqrt[4]{\dfrac{t^3+1}{t^3-1}}

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Step 1. Take the natural log of both sides: ln⁡s=14[ln⁡(t3+1)−ln⁡(t3−1)]\ln s = \dfrac14\big[\ln(t^3+1)-\ln(t^3-1)\big].

Step 2. Differentiate implicitly (chain rule on the left, chain rule on each log on the right): s′s=14[3t2t3+1−3t2t3−1]\dfrac{s'}{s}=\dfrac14\left[\dfrac{3t^2}{t^3+1}-\dfrac{3t^2}{t^3-1}\right].

Step 3. Combine the bracket over a common denominator: 3t2t3+1−3t2t3−1=3t2⋅(t3−1)−(t3+1)(t3+1)(t3−1)=3t2⋅−2t6−1=−6t2t6−1\dfrac{3t^2}{t^3+1}-\dfrac{3t^2}{t^3-1}=3t^2\cdot\dfrac{(t^3-1)-(t^3+1)}{(t^3+1)(t^3-1)}=3t^2\cdot\dfrac{-2}{t^6-1}=\dfrac{-6t^2}{t^6-1}.

Step 4. So s′s=14⋅−6t2t6−1=−3t22(t6−1)\dfrac{s'}{s}=\dfrac14\cdot\dfrac{-6t^2}{t^6-1}=\dfrac{-3t^2}{2(t^6-1)}. …

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