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. \(\frac{\text { Function }}{y=2 \tan ^{3} x} \quad \frac{\text { Point }}{\left(\frac{\pi}{4}, 2\right)}\)

Short answer

The equation of the tangent line to the graph of the function at the point \(\left(\frac{\pi}{4}, 2\right)\) is \(y - 2 = 12(x - \frac{\pi}{4})\)

Step-by-step solution

Compute derivative of the function

The derivative of \(y = 2 \tan^3 x\) is calculated using chain rule. First, differentiate \(2 \tan^3 x\) with respect to \(\tan x\) to get \(6 \tan^2 x\) and then differentiate \(\tan x\) with respect to \(x\) to get sec^2 \(x\). The total derivative is \( y' = 6 \tan^2 x \cdot \sec^2 x \)

Substitute the point into the derivative

Substitute the point \((\frac{\pi}{4}, 2)\) into the derivative. Because \(\tan(\frac{\pi}{4}) = 1\) and \(\sec(\frac{\pi}{4}) = \sqrt{2}\), \( y'(\frac{\pi}{4}) = 6 \cdot (1^2) \cdot (2) = 12 \)

Write equation of the tangent line

Use the point-slope form of the line equation: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = \left(\frac{\pi}{4}, 2\right)\) and \(m = 12\), to write the equation of the tangent line. So, the equation becomes \(y - 2 = 12(x - \frac{\pi}{4})\)