Question

Y=tan^-1(8x\1-15x2) find dy\dx

Answer 1

Step-by-step explanation:

solution is in the attached pic !

Answer 2

Answer:

$\dfrac{dy}{ dx}=\dfrac{3}{ 1+9x^2}+\dfrac{5}{ 1+25x^2}$

Step-by-step explanation:

Given the trigonometric relation,

$y=tan^{-1}\bigg(\dfrac{8x}{1-15x^2}\bigg)$

which can be written as

$y=tan^{-1}\bigg(\dfrac{3x+5x}{1-(3x)(5x)}\bigg)$

Using the trigonometric condition,

$tan^{-1}x+tan^{-1}y=tan^{-1}\bigg(\dfrac{x+y}{1-xy}\bigg)$

we can write

$tan^{-1}\bigg(\dfrac{3x+5x}{1-(3x)(5x)}\bigg)=tan^{-1}3x+tan^{-1}5x$

$\implies y=tan^{-1}3x+tan^{-1}5x$

Differentiating with respect to $x$,

$\dfrac{dy}{dx} =\dfrac{d}{dx}\Big[tan^{-1}3x+tan^{-1}5x\Big]$

     $=\dfrac{d}{dx}tan^{-1}3x+\dfrac{d}{dx}tan^{-1}5x$

Using the trigonometric relation,

$tan^{-1}x=\dfrac{1}{x^{2} }$

we get

$\dfrac{dy}{dx}=\dfrac{1}{ 1+(3x)^2}.\dfrac{d}{dx}(3x)+\dfrac{1}{ 1+(5x)^2}.\dfrac{d}{dx}(5x)$

    $=\dfrac{1}{ 1+(3x)^2}.3+\dfrac{1}{ 1+(5x)^2}.5$

    $=\dfrac{3}{ 1+9x^2}+\dfrac{5}{ 1+25x^2}$

Therefore, $\dfrac{dy}{ dx}=\dfrac{3}{ 1+9x^2}+\dfrac{5}{ 1+25x^2}$