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

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Answer 1 · 2015-10-14T01:16:51

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

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

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

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

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

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