Question

Find the derivative of the function:

26. \(f\left( t \right) = {2^{{t^3}}}\)

Short answer

The derivative of \(f\left( t \right)\) is \(3\left( {\ln 2} \right){t^2}{2^{{t^3}}}\).

Step-by-step solution

The chain rule

If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g\left( x \right)\), then the composite function \(F = f \circ g\) defined by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) is differentiable at \(x\) and \(F'\) is given by the product \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\). In Leibniz notation, if \(y = f\left( u \right)\) and \(u = g\left( x \right)\) are both differentiable functions, then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\).

The derivative of the given function

Differentiate \(f\left( t \right)\) with respect to \(t\) as follows:

\(\begin{aligned}f'\left( t \right) &= \frac{d}{{dt}}\left( {{2^{{t^3}}}} \right)\\ &= {2^{{t^3}}}\left( {\ln 2} \right)\frac{d}{{dt}}\left( {{t^3}} \right)\\ &= {2^{{t^3}}}\left( {\ln 2} \right)\left( {3{t^2}} \right)\\ &= 3\left( {\ln 2} \right)\left( {{t^2}{2^{{t^3}}}} \right)\end{aligned}\)

Hence, the derivative of \(f\left( t \right)\) is \(3\left( {\ln 2} \right)\left( {{t^2}{2^{{t^3}}}} \right)\).