Question

Differentiate the following functions from first principles  y³ + 3xy² - x³ = 3

Answer 1

Let f(x,y)=y³+3xy²-x³=3. 

f(x+h,y+k)=3, where the small change h in x creates a small change k in y. 

f(x+h,y+k)=(y+3y²k+3yk²+k³)+3(x+h)(y²+2yk+k²)-(x³+3x²h+3xh²+h³)=3.

If h and k are sufficiently small we can ignore as insignificant terms containing h2, k2 and any terms involving the product of h and k.

f(x+h,y+k)=y³+3y²k+3xy²+6xyk+3hy²-x³-3x²h=3.

The differential is f(x+h,y+k)-f(x,y)=0:

3y²k+6xyk+3hy²-3x²h=0.

Divide through by 3h:

y²k/h+2xyk/h+y²-x²=0,

(y²+2xy)(k/h)=x²-y², and (k/h)=(x²-y²)/(y²+2xy).

Limit as h,k→0 makes k/h=dy/dx=(x²-y²)/(y²+2xy).