Question

1-22: Differentiate.

6. \(g\left( x \right) = 3x + {x^2}\cos x\)

Short answer

The required value is \(g'\left( x \right) = 3 - {x^2}\sin x + 2x\cos x\).

Step-by-step solution

 Step 1: Write the formula of the derivatives of trigonometric functions, the product rule and the power function

\(\frac{d}{{dx}}\left( {\cos x} \right) = - \sin x\)

\(\frac{d}{{dx}}\left( {\sin x} \right) = \cos x\)

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

The Power Rule: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)where \(n\) is a positive integer

Find the differentiation of the function

Consider the function \(g\left( x \right) = 3x + {x^2}\cos x\). Differentiate the function w.r.t \(x\) by using the derivatives of trigonometric functions, the product rule and the power function.

\(\begin{aligned}\frac{{d\left( {g\left( x \right)} \right)}}{{dx}} &= \frac{{d\left( {3x + {x^2}\cos x} \right)}}{{dx}}\\ &= \frac{d}{{dx}}\left( {3x} \right) + {x^2}\frac{d}{{dx}}\left( {\cos x} \right) + \cos x\frac{d}{{dx}}\left( {{x^2}} \right)\\ &= 3\left( 1 \right) + {x^2}\left( { - \sin x} \right) + \cos x\left( {2x} \right)\\ &= 3 - {x^2}\sin x + 2x\cos x\end{aligned}\)

Thus, the derivative of the function \(g\left( x \right) = 3x + {x^2}\cos x\) is \(g'\left( x \right) = 3 - {x^2}\sin x + 2x\cos x\).