How do you take the derivative of ${(tan (8 x))}^{2}$ $(tan(8x))^2$ ?

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Answer 1 · 2015-06-17T13:28:25

$\frac{d y}{d x} = 16 tan (8 x) {sec}^{2} (8 x)$ $dy/dx=16tan(8x)sec^2(8x)$

Use the Chain Rule a couple times, along with the fact that $\frac{d}{d x} (tan (x)) = {sec}^{2} (x)$ $d/dx(tan(x))=sec^2(x)$ :

$\frac{d}{d x} ({tan}^{2} (8 x)) = 2 tan (8 x) \cdot \frac{d}{d x} (tan (8 x))$ $d/dx(tan^2(8x))=2tan(8x)*d/dx(tan(8x))$

$= 2 tan (8 x) \cdot {sec}^{2} (8 x) \cdot \frac{d}{d x} (8 x)$ $=2tan(8x)*sec^2(8x)*d/dx(8x)$

$= 16 tan (8 x) {sec}^{2} (8 x)$ $=16tan(8x)sec^2(8x)$

How do you take the derivative of (tan(8x))2(tan(8x))^2?