What is the derivative of (1/(x^2-9/(x^4))(x+7x^3)) ?

The derivative of (1/(x^2-9/(x^4))(x+7x^3)) is (62)/(x^2)+(27)/(x^4)+7

What is the first derivative of (1/(x^2-9/(x^4))(x+7x^3)) ?

The first derivative of (1/(x^2-9/(x^4))(x+7x^3)) is (62)/(x^2)+(27)/(x^4)+7

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\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) (x + 7 x^{3}))

\frac{62}{x ^{2}} + \frac{27}{x ^{4}} + 7

Solution steps

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) (x + 7 x^{3}))

Apply the Product Rule:

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

f = \frac{1}{x ^{2}} - \frac{9}{x ^{4}}, g = x + 7 x^{3}

= \frac{d}{d x} (\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) (x + 7 x^{3}) + \frac{d}{d x} (x + 7 x^{3}) (\frac{1}{x ^{2}} - \frac{9}{x ^{4}})

\frac{d}{d x} (\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) = - \frac{2}{x ^{3}} + \frac{36}{x ^{5}}

\frac{d}{d x} (x + 7 x^{3}) = 1 + 21 x^{2}

= (- \frac{2}{x ^{3}} + \frac{36}{x ^{5}}) (x + 7 x^{3}) + (1 + 21 x^{2}) (\frac{1}{x ^{2}} - \frac{9}{x ^{4}})

Simplify

(- \frac{2}{x ^{3}} + \frac{36}{x ^{5}}) (x + 7 x^{3}) + (1 + 21 x^{2}) (\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) : \frac{62}{x ^{2}} + \frac{27}{x ^{4}} + 7

= \frac{62}{x ^{2}} + \frac{27}{x ^{4}} + 7

\int_{- 1}^{1} x^{2} - 2 x + 3 d x

y = (2 y^{4} + 2 x) \frac{d y}{d x}

\int_{0}^{\frac{π}{4}} \frac{1}{cos ( x )} d x

\frac{\partial}{\partial x} (cos (x^{3}))

\frac{d}{d x} (cos (x^{2}) \cdot 2 x)

What is the derivative of (1/(x^2-9/(x^4))(x+7x^3)) ?
The derivative of (1/(x^2-9/(x^4))(x+7x^3)) is (62)/(x^2)+(27)/(x^4)+7
What is the first derivative of (1/(x^2-9/(x^4))(x+7x^3)) ?
The first derivative of (1/(x^2-9/(x^4))(x+7x^3)) is (62)/(x^2)+(27)/(x^4)+7