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

The derivative of '(1/(x^2)-9/(x^4))(x+5x^3) is '((44)/(x^2)+(27)/(x^4)+5)

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

The first derivative of '(1/(x^2)-9/(x^4))(x+5x^3) is '((44)/(x^2)+(27)/(x^4)+5)

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derivative of

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

' (\frac{44}{x ^{2}} + \frac{27}{x ^{4}} + 5)

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

=^{^{'}} (\frac{d}{d x} (\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) (x + 5 x^{3}) + \frac{d}{d x} (x + 5 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 + 5 x^{3}) = 1 + 15 x^{2}

=^{^{'}} ((- \frac{2}{x ^{3}} + \frac{36}{x ^{5}}) (x + 5 x^{3}) + (1 + 15 x^{2}) (\frac{1}{x ^{2}} - \frac{9}{x ^{4}}))

Expand

(- \frac{2}{x ^{3}} + \frac{36}{x ^{5}}) (x + 5 x^{3}) + (1 + 15 x^{2}) (\frac{1}{x ^{2}} - \frac{9}{x ^{4}}) : \frac{44}{x ^{2}} + \frac{27}{x ^{4}} + 5

=^{'} (\frac{44}{x ^{2}} + \frac{27}{x ^{4}} + 5)

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x \to - 1 lim (\frac{x ^{2}}{x + 1})

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\int \frac{x ^{2}}{x ^{2} + 5} d x

What is the derivative of '(1/(x^2)-9/(x^4))(x+5x^3) ?
The derivative of '(1/(x^2)-9/(x^4))(x+5x^3) is '((44)/(x^2)+(27)/(x^4)+5)
What is the first derivative of '(1/(x^2)-9/(x^4))(x+5x^3) ?
The first derivative of '(1/(x^2)-9/(x^4))(x+5x^3) is '((44)/(x^2)+(27)/(x^4)+5)