Question

Find the derivative of the function.

42. \(y = {e^{{\rm{sin}}2x}} + {\rm{sin}}\left( {{e^{2x}}} \right)\)

Short answer

The derivative of the function is \(2{e^{{\rm{sin}}2x}}{\rm{cos}}2x + 2{\rm{cos}}\left( {{e^{2x}}} \right){e^{2x}}\), that is, \(y' = 2{e^{{\rm{sin}}2x}}{\rm{cos}}2x + 2{\rm{cos}}\left( {{e^{2x}}} \right){e^{2x}}\).

Step-by-step solution

Sum rule of differentiation

Use sum rule of differentiation which implies that \(\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\).

Apply sum rule;

\(\begin{aligned}y' &= \frac{d}{{dx}}\left( {{e^{{\rm{sin}}2x}} + {\rm{sin}}\left( {{e^{2x}}} \right)} \right)\\& = \frac{d}{{dx}}\left( {{e^{{\rm{sin}}2x}}} \right) + \frac{d}{{dx}}\left( {{\rm{sin}}\left( {{e^{2x}}} \right)} \right)\end{aligned}\)

Chain rule

According to the chain rule, \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\), where \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) and \(f\) is differentiable at \(g\left( x \right)\) .

Find the derivative of \(y\) with respect to \(x\).

\(\begin{aligned}y' &= \frac{d}{{dx}}\left( {{e^{{\rm{sin}}2x}}} \right) + \frac{d}{{dx}}\left( {{\rm{sin}}\left( {{e^{2x}}} \right)} \right)\\ &= {e^{\sin 2x}}\frac{d}{{dx}}\left( {{\rm{sin}}2x} \right) + {\rm{cos}}\left( {{e^{2x}}} \right)\frac{d}{{dx}}\left( {{e^{2x}}} \right)\\ &= {e^{{\rm{sin}}2x}}{\rm{cos}}2x\frac{d}{{dx}}\left( {2x} \right) + {\rm{cos}}\left( {{e^{2x}}} \right){e^{2x}}\frac{d}{{dx}}\left( {2x} \right)\\ &= {e^{{\rm{sin}}2x}}{\rm{cos}}2x\left( 2 \right) + {\rm{cos}}\left( {{e^{2x}}} \right){e^{2x}}\left( 2 \right)\\ &= 2{e^{{\rm{sin}}2x}}{\rm{cos}}2x + 2{\rm{cos}}\left( {{e^{2x}}} \right){e^{2x}}\end{aligned}\)

Thus, the derivative of the function is \(2{e^{{\rm{sin}}2x}}{\rm{cos}}2x + 2{\rm{cos}}\left( {{e^{2x}}} \right){e^{2x}}\), that is, \(y' = 2{e^{{\rm{sin}}2x}}{\rm{cos}}2x + 2{\rm{cos}}\left( {{e^{2x}}} \right){e^{2x}}\).