Section 3.9 : Chain Rule
17. Differentiate \(h\left( t \right) = {t^6}\,\sqrt {5{t^2} - t} \) .
Show Solution
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,
\[\begin{align*}h\left( t \right) & = {t^6}\,{\left( {5{t^2} - t} \right)^{\frac{1}{2}}}\\ h'\left( t \right) & = 6{t^5}{\left( {5{t^2} - t} \right)^{\frac{1}{2}}} + {t^6}\left( {\frac{1}{2}} \right){\left( {5{t^2} - t} \right)^{ - \,\,\frac{1}{2}}}\left( {10t - 1} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{6{t^5}{{\left( {5{t^2} - t} \right)}^{\frac{1}{2}}} + \frac{1}{2}{t^6}\left( {10t - 1} \right){{\left( {5{t^2} - t} \right)}^{ - \,\,\frac{1}{2}}}}}\end{align*}\]