Find the equation of the tangent to the curve y^3 - 4x^2 - 3xy + 25 = 0 at the point (2,-3).

This is an example of implicit differentiation where we have to consider both the y-terms as well as the x-terms. The first step is to deal with each term on the left hand side individually. This gives us 3y^2(dy/dx) - 8x for the first two terms. The '-3xy' however, requires the product rule which is v(du/dx) + u(dv/dx) in order for it to be differentiated. Using -3x as our value for u and y as our value for v giving us -3y-3x(dx/dy) when differentiated. The last term differentiates to give zero. Putting this altogether gives us: 3y^2(dy/dx) - 8x - 3y - 3x(dy/dx) = 0.The next step is to separate the (dy/dx) terms from the other terms by factorising and then taking the other terms to the other side giving us (dy/dx)(3y^2-3x)=8x+3y.We then rearrange to get (dy/dx) by itself in order to find the gradient using (dy/dx)=(8x+3y)/(3y^2-3x).We plug in the given values for the x and y-coordinates into the expression to find that the gradient is 1/3 which will also be the gradient of the tangent at that point. Using this in the equation for a straight line, which we are expected to know: y-y1=m(x-x1) where m is the gradient. This gives us y+3=(1/3)(x-2) as the equation of the tangent.