Question

find dy/dx in the following : cos(x+y)+y^2=3​

Answer 1

Answer:

Use the formulas like....

uv= udv/dx + vdu/dx ( product rule)

sin(x+y)= sinxcosy+cosxsiny.

.....

hope it'll be helpful for you

Answer 2

Answer:

The derivative is  

$\dfrac{dy}{dx} = \dfrac{\sin(x+y)}{2y-\sin(x+y)}$

Step-by-step explanation:

Given the expression,

$\cos(x+y)+y^2=3$

Since the function contains both $x$ and $y$, we need to perform implicit differentiation to get the derivative $\frac{dy}{dx}$ .

That is we should differentiate the whole expression with respect to $x$ .

We know that

$\frac{d}{dx}\cos(x)=-\sin (x)$

and $\frac{d}{dx}x^2= 2x$

Thus we have,

$\dfrac{d}{dx}(\cos(x+y)+y^2)=\dfrac{d}{dx}(3) \\\\\implies \dfrac{d}{dx}\cos(x+y)+\dfrac{d}{dx}y^2=0\\\\\implies -\sin(x+y)\dfrac{d}{dx}(x+y)+2y\dfrac{dy}{dx}=0\\\\\implies -\sin(x+y)(1+\dfrac{dy}{dx})+2y\dfrac{dy}{dx}=0\\\\\implies -\sin(x+y)+(-\sin(x+y)\frac{dy}{dx}+2y\dfrac{dy}{dx}=0\\\\\implies \dfrac{dy}{dx}(2y-\sin(x+y))=\sin(x+y)\\\\\implies \dfrac{dy}{dx} = \dfrac{\sin(x+y)}{2y-\sin(x+y)}$