Find the gradient of y^2 +2xln(y) = x^2 at the point (1,1)
The key point in this question is noting that implicit differentiation is a very useful tool, although it is slightly different and looks much scarier than differentiating y = f(x).To do the question, a mix of the chain rule (and the product rule for the 2xln(y) term in the middle) is sufficient to differentiate this, and then it is simply a case of rearranging and substituting the value in. y2+2xln(y) = x2=> 2(dy/dx)y + 2lny + (2x/y)(dy/dx) = 2x=> dy/dx = (x-lny)/(y + x/y)So when x=y=1, dy/dx = 1/2
