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Section 3.9 : Chain Rule
17. Differentiate \(h\left( t \right) = {t^6}\,\sqrt {5{t^2} - t} \) .
Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of the terms in the product or quotient.
For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll need to use the Chain Rule when we differentiate the second term.
The derivative is then,
\[\begin{align*}h\left( t \right) & = {t^6}\,{\left( {5{t^2} - t} \right)^{\frac{1}{2}}}\\ h'\left( t \right) & = 6{t^5}{\left( {5{t^2} - t} \right)^{\frac{1}{2}}} + {t^6}\left( {\frac{1}{2}} \right){\left( {5{t^2} - t} \right)^{ - \,\,\frac{1}{2}}}\left( {10t - 1} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{6{t^5}{{\left( {5{t^2} - t} \right)}^{\frac{1}{2}}} + \frac{1}{2}{t^6}\left( {10t - 1} \right){{\left( {5{t^2} - t} \right)}^{ - \,\,\frac{1}{2}}}}}\end{align*}\]