Derivation Quotient Rule 5x^8-3/3x-7?

Question

I get dy/dx = [(3x - 7) (40x^7) - (5x^8 - 3) (3)] / (3x - 7)^2

But how to simplify it further? Can show me the steps and explain? Thanks

Answer 1

y = (5x^8 - 3) / (3x - 7)

dy/dx = [(3x - 7) (40x^7) - (5x^8 - 3) (3)] / (3x - 7)^2

We are going to break up fraction with the form:
(a - b) / c = a/c - b/c

dy/dx = [(3x - 7) (40x^7) - (5x^8 - 3) (3)] / (3x - 7)^2
dy/dx = (3x - 7) (40x^7) / (3x - 7)^2 - (5x^8 - 3) (3) / (3x - 7)^2

Cancel (3x - 7) in first term:
dy/dx = (40x^7) / (3x - 7) - (5x^8 - 3) (3) / (3x - 7)^2

Rename second term:
dy/dx = (40x^7) / (3x - 7) - A

We can further simplify by substituting y = (5x^8 - 3) / (3x - 7)

A = (5x^8 - 3) (3) / (3x - 7)^2
A = [ (5x^8 - 3) / (3x - 7) ] * [ 3 / (3x - 7) ]
A = y * [ 3 / (3x - 7) ]
A = 3y / (3x - 7)

Then,
dy/dx = (40x^7) / (3x - 7) - A
dy/dx = (40x^7) / (3x - 7) - 3y / (3x - 7)
dy/dx = (40x^7 - 3y) / (3x - 7)

Answer 2

Assuming that question should be shown as :-
f(x) = (5x^8 - 3) / (3x - 7)
f `(x)
= [(3x - 7)(40x^7) - (5x^8 - 3)(3) ] / (3x - 7)²
= [ 120 x^8 - 280x^7 - 15x^8 + 9 ] / (3x - 7)²
= [ 105 x^8 - 280x^7 + 9 ] / (3x - 7)²