Question

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=\cos 3 x} \quad \frac{\text { Point }}{\left(\frac{\pi}{4},-\frac{\sqrt{2}}{2}\right)}\)

Short answer

The equation of the tangent line to the graph of \(y=\cos3x\) at the point \(\frac{\pi}{4}, -\frac{\sqrt{2}}{2}\) is \(y = -3\frac{\sqrt{2}}{2}x + \frac{3\pi}{4\sqrt{2}}\)

Step-by-step solution

Calculate derivative of the function

The derivative of \(y=\cos3x\) is calculated using chain rule of derivatives which results in \(-3\sin3x\).

Evaluate derivative at the given point

Evaluate the derivative at the point \(\frac{\pi}{4}\). This gives the slope of the tangent line at the given point. \[-3\sin3(\frac{\pi}{4}) = -3\sin\frac{3\pi}{4} = -3(\frac{\sqrt{2}}{2}) = -3\frac{\sqrt{2}}{2}\]. Therefore, the slope of the tangent line at the point \(\frac{\pi}{4}, -\frac{\sqrt{2}}{2}\) is \(-3\frac{\sqrt{2}}{2}\).

Find the equation of the tangent line

Use the slope-point form of a line, which is \(y-y_1 = m(x-x_1)\), where \((x_1,y_1)\) is the point of tangency and \(m\) is the slope of the tangent. Here, \(x_1 = \frac{\pi}{4}\), \(y_1 = -\frac{\sqrt{2}}{2}\), and \(m = -3\frac{\sqrt{2}}{2}\), the equation becomes \(y + \frac{\sqrt{2}}{2} = -3\frac{\sqrt{2}}{2}(x - \frac{\pi}{4})\)

Simplify the equation of the tangent line

Simplify the equation obtained in Step 3 to get the final equation: \(y = -3\frac{\sqrt{2}}{2}x + \frac{3\pi}{4\sqrt{2}}\)