Find the derivative of y = 2sqrt(3 arc tan 5x)?

Question

Find the derivative of y = 2sqrt(3 arc tan 5x)

Answer 1

y = 2( 3tan-15x)^-1/2
let, u = 3tan^-1( 5x)

now the next series of processes in are to show u how to differential the inverse of a tan function.
Let y= tan^-1(x)
=> tan y = x
differentiating we av,
sec^2(y) dy/dx = 1
=> dy/dx= 1/sec^2(y)
but, 1 + tan^2(y) = sec^2(y)

=> dy/dx= 1/ (1 + tan^2(y) )
but, x = tan y
=> dy/dx = 1/ (1 + x^2)
so lets go back to the question.
From above explanation,
=> du/dx = 15/ ( 1 + 25x^2)

y = 2u^(1/2)
since, u = 3tan^-1 (5x) and y = 2( 3tan-15x)^(-1/2)

therefore, dy/du = u^(-1/2)
therefore, dy/dx = dy/du * du/dx
dy/dx = ( u^(-1/2) ) * (15/ (1 + 25x^2) )
but, u = 3tan^-1 (5x)
dy/dx = (1 / (3tan-15x)^(1/2)) * ( 15/ (1 + 25x^2) )

therefore, dy/dx = 15 / ( (1 + 25x^2) * (3tan-15x)^(1/2) )

Answer 2

y = 2sqrt(3 arc tan 5x)

let p = 3 tan^-1 (5x) >>>> dp/dx = 3 [1/(1+25x^2)]. 5
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d/dx [ acr tan x] = 1/(1+x^2) and 5 multiplied bcause differ of 5x
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dp/dx = 15 /[1+25x^2].

y = 2 sqrt (p) >>> dy/dp = 2 (1/2) (p)^-1/2

dy/dx = dy/dp * dp/dx

dy/dx = (p)^-1/2 *15 /[1+25x^2].

dy/dx = 15 (3 tan^-1 (5x))^-1/2 / [1+25x^2]

dy/dx = 15 / { [sqrt (3 tan^-1 (5x)) ] * (1+25x^2) }