Differentiate the following functions from first principles  y³ + 3xy² - x³ = 3
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Consider a point (x+h,y+k) where h and k are tiny displacements from x and y, that is, this point is very close to (x,y). The displacements are so tiny that we can ignore all second and more degree quantities: terms containing h², k² and hk and anything with a higher degree. Plug the point into the equation and we get the approximation: (y³+3ky²)+3(x+h)(y²+2ky)-(x³+3hx²)=3.

This expands to y³+3ky²+3xy²+6kxy+3hy²-x³-3hx²=3.

From this we subtract the equation involving just x and y:

3ky²+6kxy+3hy²-3hx²=0, ky²+2kxy+hy²-hx²=0.

Divide through by h:

(k/h)y²+2(k/h)xy+y²-x²=0.

k/h=(x²-y²)/(y²+2xy) is the gradient between (x,y) and (x+h,y+k) and is the definition of differentiation.

So the differential at the point (x,y) is (x²-y²)/(y²+2xy).

by Top Rated User (660k points)

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