What is the derivative of ${tan}^{5} (x)$ $tan^5(x)$ ?

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What is the derivative of ${tan}^{5} (x)$ $tan^5(x)$ ?

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Answer 1 · 2018-03-21T18:03:38

$y = {tan}^{5} x = {(tan x)}^{5} \Rightarrow \frac{d y}{d x} = 5 {(tan x)}^{4} \frac{d}{d x} (tan x)$ $y=tan^5x=(tanx)^5=>dy/dx=5(tanx)^4d/(dx)(tanx)$
$\Rightarrow \frac{d y}{d x} = 5 {tan}^{4} x \cdot {sec}^{2} x$ $=>dy/dx=5tan^4x*sec^2x$

Explanation:

Let,
$y = {tan}^{5} x = {(tan x)}^{5}$ $y=tan^5x=(tanx)^5$
We take,
$y = u^{5},$ $y=u^5,$ where, $u = tan x$ $u=tanx$
$\Rightarrow \frac{d y}{d u} = 5 u^{4} and \frac{d u}{d x} = {sec}^{2} x$ $=>(dy)/(du)=5u^4 and (du)/(dx)=sec^2x$
Applying chain rule
$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$
$\Rightarrow \frac{d y}{d x} = 5 u^{4} \cdot {sec}^{2} x$ $=>dy/dx=5u^4*sec^2x$
$\Rightarrow \frac{d y}{d x} = 5 {(tan x)}^{4} \cdot {sec}^{2} x$ $=>dy/dx=5(tanx)^4*sec^2x$
$\Rightarrow \frac{d y}{d x} = 5 {tan}^{4} x {sec}^{2} x$ $=>dy/dx=5tan^4xsec^2x$

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