Question

Find the derivative of the function.

33. \(F\left( t \right) = {e^{t\sin 2t}}\)

Short answer

The derivative of \(F\left( t \right)\) is \({e^{t\sin 2t}}\left( {2t\cos 2t + \sin 2t} \right)\).

Step-by-step solution

The product rule of differentiation

If a function \(h\left( x \right) = f\left( x \right) \cdot g\left( x \right)\) and \(f\), \(g\) are both differentiable, then the derivative of \(h\left( x \right)\) is as follows:

\(\frac{{dh}}{{dx}} = \frac{d}{{dx}}\left( {f\left( x \right) \cdot g\left( x \right)} \right) = f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

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:

\(\begin{aligned}F'\left( t \right) &= \frac{d}{{dt}}\left( {{e^{t\sin 2t}}} \right)\\ &= {e^{t\sin 2t}}\frac{d}{{dt}}\left( {t\sin 2t} \right)\\ &= {e^{t\sin 2t}}\left( {t\frac{d}{{dt}}\left( {\sin 2t} \right) + \left( {\sin 2t} \right)\frac{d}{{dt}}\left( t \right)} \right)\\ &= {e^{t\sin 2t}}\left( {t\left( {\cos 2t} \right)\frac{d}{{dt}}\left( {2t} \right) + \left( {\sin 2t} \right)\left( 1 \right)} \right)\\ &= {e^{t\sin 2t}}\left( {t\left( {\cos 2t} \right)\left( 2 \right) + \sin 2t} \right)\\ &= {e^{t\sin 2t}}\left( {2t\cos t + \sin 2t} \right)\end{aligned}\)

Hence, the derivative of \(F\left( t \right)\) is \({e^{t\sin 2t}}\left( {2t\cos 2t + \sin 2t} \right)\).