Given:
The equation is
[tex]y=\cot(x+y)[/tex]
To find:
The value of [tex]\dfrac{dy}{dx}[/tex].
Solution:
We have,
[tex]y=\cot(x+y)[/tex]
Differentiate with respect to x.
[tex]\dfrac{dy}{dx}=\dfrac{d}{dx}\cot(x+y)[/tex]
[tex]\dfrac{dy}{dx}=-\text{cosec}^2(x+y)\dfrac{d}{dx}(x+y)[/tex]
[tex]\dfrac{dy}{dx}=-\text{cosec}^2(x+y)(1+\dfrac{dy}{dx})[/tex]
[tex]\dfrac{dy}{dx}=-\text{cosec}^2(x+y)-\text{cosec}^2(x+y)\dfrac{dy}{dx}[/tex]
[tex]\dfrac{dy}{dx}+\text{cosec}^2(x+y)\dfrac{dy}{dx}=-\text{cosec}^2(x+y)[/tex]
[tex]\dfrac{dy}{dx}(1+\text{cosec}^2(x+y))=-\text{cosec}^2(x+y)[/tex]
[tex]\dfrac{dy}{dx}=-\dfrac{\text{cosec}^2(x+y)}{1+\text{cosec}^2(x+y)}[/tex]
Therefore, [tex]\dfrac{dy}{dx}=-\dfrac{\text{cosec}^2(x+y)}{1+\text{cosec}^2(x+y)}[/tex].
