What is the derivative of ${tan}^{7} (x^{2})$ $tan^7(x^2)$ ?

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Answer 1 · 2017-08-14T21:22:12

$\frac{d}{d x} [{tan}^{7} (x^{2})] = 14 x {tan}^{6} (x^{2}) {sec}^{2} (x^{2})$ $d/(dx) [tan^7(x^2)] = color(blue)(14xtan^6(x^2)sec^2(x^2)$

We're asked to find the derivative

$\frac{d}{d x} [{tan}^{7} (x^{2})]$ $d/(dx) [tan^7(x^2)]$

We can first use the chain rule:

$\frac{d}{d x} [{tan}^{7} (x^{2})] = \frac{d}{d u} [u^{7}] \frac{d u}{d x}$ $d/(dx) [tan^7(x^2)] = d/(du) [u^7] (du)/(dx)$

where

$= 7 \frac{d}{d x} [tan (x^{2})] {tan}^{6} (x^{2})$ $= 7d/(dx)[tan(x^2)]tan^6(x^2)$

Using the chain rule again:

$\frac{d}{d x} [tan (x^{2})] = \frac{d}{d u} [tan u] \frac{d u}{d x}$ $d/(dx) [tan(x^2)] = d/(du) [tanu] (du)/(dx)$

where

$= 7 \frac{d}{d x} [x^{2}] {sec}^{2} (x^{2}) {tan}^{6} (x^{2})$ $= 7d/(dx)[x^2]sec^2(x^2)tan^6(x^2)$

Use the power rule on the $x^{2}$ $x^2$ term:

$\frac{d}{d x} [x^{n}] = n x^{n - 1}$ $d/(dx) [x^n] = nx^(n-1)$

where

$= 7 (2 x) {sec}^{2} (x^{2}) {tan}^{6} (x^{2})$ $= 7(2x)sec^2(x^2)tan^6(x^2)$

$color(blue)(ulbar(|stackrel(" ")(" "= 14xsec^2(x^2)tan^6(x^2)" ")|)$

What is the derivative of tan^7(x^2)tan7(x2)tan^7(x^2)?