Question

how to find the derivative of xy = In (x² + y ² )? please show all workings and simplify!


the answer is supposed to be yx^2+y^3 -2x / 2y - x^3-xy^2

Answer 1

Answer:

Step-by-step explanation:

Keep in mind that the derivative of ln(u) = u'/u.  Here, our u is x^2 + y^2, so the derivative of that will fit in for u'.  Let's do this step by step:

$xy=ln(x^2+y^2)$

Working on the left first, using the product rule, the derivative (implicite, of course!) is:

$x\frac{dy}{dx}+1y=ln(x^2+y^2)$

Now we will work on the right side, keeping in mind the rule above for derivatives of natural logs:

$x\frac{dy}{dx}+1y=\frac{2x+2y\frac{dy}{dx} }{x^2+y^2}$

Now we are going to get rid of the donominator on the right by multiplication on both sides:

$(x^2+y^2)(x\frac{dy}{dx}+1y)=2x+2y\frac{dy}{dx}$

Distribute on the left to get

$x^3\frac{dy}{dx}+x^2y+xy^2\frac{dy}{dx}+y^3=2x+2y\frac{dy}{dx}$

Now collect all the terms with dy/dx in them on one side and everything else goes on the other side:

$x^3\frac{dy}{dx}+xy^2\frac{dy}{dx}-2y\frac{dy}{dx}=2x-x^2y-y^3$

Factor out the common dy/dx:

$\frac{dy}{dx}(x^3+xy^2-2y)=2x-x^2y-y^3$

and divide on the left to isolate the dy/dx:

$\frac{dy}{dx}=\frac{2x-x^2y-y^3}{x^3+xy^2-2y}$

And there you go!