Using yours, one point would be (k, k³ + 12k - 9) and the other would be (k + h, [k+h]³ + 12[k + h] - 9). {Different brackets used for clarity.}

Expanding the y-part of k+h gives k³ + 3k²h + 3kh² + h³ + 12k + 12h - 9.

That means that the difference in y is 3k²h + 3kh² + h³ + 12h and the difference in x is h, obviously.

The gradient is found by dividing the difference in y by the difference in x, which is 3k² + 3kh + h² + 12. The gradient at a point would be when k and k+h are in the same place and so h would be 0. Substituting h = 0 into the gradient gives dy/dx = 3k² + 12. Put x back in to get 3x² + 12.

]]>How can x=k and x=k+h at the same time?

anyway, what do you want to diferentiate? y(x)?

Do you want to find DY? DX? Or maybe DY/DX???????????????

Please restate your question in a more clear way

]]>Doing x = k and x = k+h I am doing differentiation

Anyway you get:

k^3 + 3k^2h + 3kh^2 + h^3 + 12k + 12h - 9 - (k^3+12k-9)

over k+h-k

Anyway I cancel the k's at the bottom and then divide the top by h

Now the answer states

3k^2h + h^3 +12h

Over h

I dont get how they got rid of the 3kh^2

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