Question

if y= tan^-1 5x/1-6x^2

Answer 1

Answer:

Step-by-step explanation:y=tan^-1[2x+3x/1-(2x)(3x)]

Let 2x be tanA and 3x be tanB

y=tan^-1[tanA - tanB/1 - tanAtanB ]

y = tan^-1 [ tan ( A + B) ]

y = A + B

y = 2x +3x

y = 5x

Derivative with respect to x:-

dy/dx = 5 Ans

Answer 2

Answer:

Required answer is 5.

Step-by-step explanation:

Given,

$y = { \tan( \frac{5x}{1 - 6 {x}^{2} } ) }^{ - 1}$

$y = { \tan( \frac{3x + 2x}{1 - 2x \times 3x} ) }^{ - 1}$

Let 2x= tanp and 3x=tanq

So,

$y = { \tan( \frac{tanq+ tanp}{1 - tanq \times tanp} ) }^{ - 1}$

$y = { \tan( \tan((p + q) ) }^{ - 1}$

$y = p + q \\ y = 2x + 3x \\ y = 5x$

Do derivative with respect to x and get,

$\frac{dy}{ dx} = 5$

Required aanswer is 5