How do you differentiate y = (2x^3 + 4 )/( x + 7)y=2x3+4x+7?

1 Answer
Oct 14, 2015

f^'(x) = (4x^3 + 42x^2 + 4)/(x+7)^2

Explanation:

If we can notate the function in the form of

f(x) = g(x)/(h(x))

We can say that

f^'(x) = (g^'(x)h(x) - g(x)h^'(x))/(h^2(x)), so

f^'(x) = ((6x^2)(x+7) - (2x^3+4)(1))/(x+7)^2

f^'(x) = (6x^3 + 42x^2 - 2x^3 + 4)/(x+7)^2

f^'(x) = (4x^3 + 42x^2 + 4)/(x+7)^2