Let y=f(x).
Let y+k=f(x+h) such that h and k are small enough to discount any 2nd degree order or higher.
y+k=7(x+h)⁵-5(x+h)³=7(x⁵+5hx⁴)-5(x³+3hx²), because we ignore terms containing h², k², and higher powers.
y+k=7x⁵+35hx⁴-5x³-15hx². y=7x⁵-5x³.
Substituting for y: k=35hx⁴-15hx². Therefore k/h=35x⁴-15x².
As h→0, k→0, and h/k→dy/dx, the derivative=35x⁴-15x².