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#