If I had an equation with both 'x' and 'y' present, how would I find the gradient?
Using the equation (x+y)2=xy2 as an example base to see how it would work in practice. The most important fact is that dy/dx =y'. Simply put, the derivative of y is dy/dx. This allows you to go through an equation using the same principles already known for x, then once again for y. This identity builds onto the principles of partial differentiation learnt in further maths and Uni level maths.
So we would go about differentiating this normally for x, then once again for y. This will produce a not so nice looking equation containing x's, y's and dy/dx's. However, you can group all the dy/dx's to one side of the equation to produce a statement for the differentiation of the equation, and hence its gradient at a nodal point.
For the example equation, the final solution would be dy/dx = (y2-2x-2y)/(2x+2y-2xy). Inputting an x value of say x=1, the gradient at that location would be -3/8.
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I already done this.
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