What is the derivative of (1/(x^2-5/(x^4))(x+9x^3)) ?

The derivative of (1/(x^2-5/(x^4))(x+9x^3)) is (44)/(x^2)+(15)/(x^4)+9

What is the first derivative of (1/(x^2-5/(x^4))(x+9x^3)) ?

The first derivative of (1/(x^2-5/(x^4))(x+9x^3)) is (44)/(x^2)+(15)/(x^4)+9

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

\frac{44}{x ^{2}} + \frac{15}{x ^{4}} + 9

Solution steps

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

Apply the Product Rule:

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

f = \frac{1}{x ^{2}} - \frac{5}{x ^{4}}, g = x + 9 x^{3}

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

\frac{d}{d x} (\frac{1}{x ^{2}} - \frac{5}{x ^{4}}) = - \frac{2}{x ^{3}} + \frac{20}{x ^{5}}

\frac{d}{d x} (x + 9 x^{3}) = 1 + 27 x^{2}

= (- \frac{2}{x ^{3}} + \frac{20}{x ^{5}}) (x + 9 x^{3}) + (1 + 27 x^{2}) (\frac{1}{x ^{2}} - \frac{5}{x ^{4}})

Simplify

(- \frac{2}{x ^{3}} + \frac{20}{x ^{5}}) (x + 9 x^{3}) + (1 + 27 x^{2}) (\frac{1}{x ^{2}} - \frac{5}{x ^{4}}) : \frac{44}{x ^{2}} + \frac{15}{x ^{4}} + 9

= \frac{44}{x ^{2}} + \frac{15}{x ^{4}} + 9

x \to 0 + lim (\frac{( 3 )}{x ^{2} - x})

\int_{0}^{8} \frac{x}{( x + 8 ) ^{3}} d x

\frac{d}{d x} ((3 x^{2} - 5 x) e^{x})

\int \frac{x}{x - 1} d x

y = \frac{e ^{x}}{5 x ^{2}}

What is the derivative of (1/(x^2-5/(x^4))(x+9x^3)) ?
The derivative of (1/(x^2-5/(x^4))(x+9x^3)) is (44)/(x^2)+(15)/(x^4)+9
What is the first derivative of (1/(x^2-5/(x^4))(x+9x^3)) ?
The first derivative of (1/(x^2-5/(x^4))(x+9x^3)) is (44)/(x^2)+(15)/(x^4)+9