Question

1-22 Differentiate.

12. \(f\left( x \right) = {e^x}\sin x + \cos x\)

Short answer

The differentiation of the function \(f\left( x \right) = {e^x}\sin x + \cos x\) is \(f'\left( x \right) = {e^x}\left( {\cos x + \sin x} \right) - \sin x\).

Step-by-step solution

Write the formula of the derivatives of trigonometric functions, exponential function and the product rule

\(\begin{aligned}\frac{d}{{dx}}\left( {\cos x} \right) &= - \sin x\\\frac{d}{{dx}}\left( {\sin x} \right) &= \cos x\\\frac{d}{{dx}}\left( {{e^x}} \right) &= {e^x}\end{aligned}\)

The product rule: If \(f\) and \(g\) are differentiable functions, then\(\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\frac{d}{{dx}}\left( {f\left( x \right)} \right)\).

Find the differentiation of the function 

Consider the function \(f\left( x \right) = {e^x}\sin x + \cos x\). Differentiate the function w.r.t \(x\) by using the derivatives of trigonometric functions, exponential function and the product rule.

\(\begin{aligned}\frac{{d\left( {f\left( x \right)} \right)}}{{dx}} &= \frac{{d\left( {{e^x}\sin x + \cos x} \right)}}{{dx}}\\ &= \frac{d}{{dx}}\left( {{e^x}\sin x} \right) + \frac{d}{{dx}}\left( {\cos x} \right)\\ &= {e^x}\frac{d}{{dx}}\left( {\sin x} \right) + \sin x\frac{d}{{dx}}\left( {{e^x}} \right) + \frac{d}{{dx}}\left( {\cos x} \right)\\ &= {e^x}\cos x + \sin x\left( {{e^x}} \right) - \sin x\\ &= {e^x}\left( {\cos x + \sin x} \right) - \sin x\end{aligned}\)

Thus, the derivative of the function \(f\left( x \right) = {e^x}\sin x + \cos x\) is \(f'\left( x \right) = {e^x}\left( {\cos x + \sin x} \right) - \sin x\).