Question

find the derivatives of root tan x . ( write properly in page , plz )
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Answer 1

Answer:

$\frac{ {( \sec(x)) }^{2} }{ \sqrt{ \tan(x) } }$

Step-by-step explanation:

My handwriting is very bad so I thought of texting the whole answer

in order to find the differential of root tan x we need to know the chain rule.

it states that

$\frac{dy}{dx} (f(g(x)) = \frac{ df}{dg} \times \frac{dg}{dx}$

Lets take an example

$\frac{d({2x + 1}^{2}) }{dx} = \frac{da}{db} \times \frac{db}{dx}$

here

$a = {b}^{2} = {(2x + 1)}^{2} \\ b = 2x + 1$

so

$\frac{da}{db} = \frac{d( {b}^{2}) }{db} = 2 {b}^{2 - 1} = 2b$

and

$\frac{db}{dx} = \frac{d}{dx} (2x) + \frac{d}{dx} (1) \\ = 2 + 0 = 2$

similarly,

$\frac{dy}{dx} = \frac{dy}{dz} \times \frac{dz}{dx}$

where

$y = \sqrt{ z } = \sqrt{ \tan(x) } \\ z = \tan(x)$

we know

$\frac{dy}{dz} = \frac{d}{dz} ( \sqrt{z} ) = \frac{d}{dz} ( {z}^{ \frac{1}{2} } ) \\ = \frac{1}{2} \times {z}^{ \frac{1}{2} - 1} = \frac{1}{2} \times {z}^{ - \frac{1}{2} } \\ = \frac{1}{ \sqrt{z} } = \frac{1}{ \sqrt{ \tan(x) } }$

and

$\frac{dz}{dx} = \frac{d}{dx} ( \tan(x)) = { \sec(x) }^{2}$

so the final answer is

$\frac{ dy }{dx} = \frac{ { (\sec(x)) }^{2} }{ \sqrt{ \tan(x) } }$

Please mark me as the brainliest