What is the derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) ?

The derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) is (12x^5-2x^3-3x-12)/(x^3)+15

What is the first derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) ?

The first derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) is (12x^5-2x^3-3x-12)/(x^3)+15

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derivative of

f (x) = (x^{3} + 3 x + 6) (4 + \frac{1}{x ^{2}})

\frac{12 x ^{5} - 2 x ^{3} - 3 x - 12}{x ^{3}} + 15

Solution steps

\frac{d}{d x} ((x^{3} + 3 x + 6) (4 + \frac{1}{x ^{2}}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3} + 3 x + 6, g = 4 + \frac{1}{x ^{2}}

= \frac{d}{d x} (x^{3} + 3 x + 6) (4 + \frac{1}{x ^{2}}) + \frac{d}{d x} (4 + \frac{1}{x ^{2}}) (x^{3} + 3 x + 6)

\frac{d}{d x} (x^{3} + 3 x + 6) = 3 x^{2} + 3

\frac{d}{d x} (4 + \frac{1}{x ^{2}}) = - \frac{2}{x ^{3}}

= (3 x^{2} + 3) (4 + \frac{1}{x ^{2}}) + (- \frac{2}{x ^{3}}) (x^{3} + 3 x + 6)

Simplify

(3 x^{2} + 3) (4 + \frac{1}{x ^{2}}) + (- \frac{2}{x ^{3}}) (x^{3} + 3 x + 6) : \frac{12 x ^{5} - 2 x ^{3} - 3 x - 12}{x ^{3}} + 15

= \frac{12 x ^{5} - 2 x ^{3} - 3 x - 12}{x ^{3}} + 15

What is the derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) ?
The derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) is (12x^5-2x^3-3x-12)/(x^3)+15
What is the first derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) ?
The first derivative of f(x)=(x^3+3x+6)(4+1/(x^2)) is (12x^5-2x^3-3x-12)/(x^3)+15