Question

Evaluate the derivative of the function at the indicated point. Use a graphing utility to verify your result. $$ f(x)=\sin x(\sin x+\cos x) \quad\left(\frac{\pi}{4}, 1\right) $$

Short answer

The derivative of the function at \(\left(\frac{\pi}{4}, 1\right)\) is \(\sqrt{2}\)

Step-by-step solution

Identify the two functions

The given function is \(f(x) = \sin x(\sin x + \cos x)\). The product rule of differentiation is applied when two functions are multiplied together. Here, you need to identify the two functions that are being multiplied. Set \(u(x) = \sin x\) and \(v(x) = \sin x + \cos x\).

Differentiate the functions

Calculate the derivative of both functions \(u(x)\) and \(v(x)\) using the rule of differentiation. The derivative of \(u(x) = \sin x\) is \(u'(x) = \cos x\). The derivative of \(v(x) = \sin x + \cos x\) is \(v'(x) = \cos x - \sin x\).

Apply the Product Rule

Now you can apply the product rule. According to the product rule of differentiation, \(d(u.v) = udv + vdu\). Therefore, differentiating \(f(x) = u(x)v(x)\) gives \(f'(x) = u'(x)v(x) + u(x)v'(x) = \cos x (\sin x + \cos x) + \sin x (\cos x - \sin x) = 2\cos{x}\sin{x}\).

Substitute the given point into the derivative

To find the value of the derivative at \((\frac{\pi}{4}, 1)\), you can substitute \(x=\frac{\pi}{4}\) into \(f'(x)\), giving \(f'\left(\frac{\pi}{4}\right) = 2 \cos \left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}\right) = \sqrt{2}\).

Verification using a graphing utility

A graphing utility can be used to confirm these results. The utility would show that the gradient at \((\frac{\pi}{4}, 1)\) is indeed \(\sqrt{2}\). Although this step is not shown here, this can be easily done on paper, on a graphing calculator or online graphing tool.