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

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

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

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

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

- \frac{4}{x ^{5}} + \frac{6}{x ^{3}} + 3 x^{2}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (\frac{1}{x ^{4}}) = - \frac{4}{x ^{5}}

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

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

= - \frac{4}{x ^{5}} - (- \frac{6}{x ^{3}}) + 3 x^{2}

Simplify

= - \frac{4}{x ^{5}} + \frac{6}{x ^{3}} + 3 x^{2}

\int \frac{1}{2 x ^{2} + 3} d x

f (x) = x^{2} - 6 x + 9, (- 2, 25)

v \to 3 lim (\frac{v + 1 - 2}{v - 3})

x \to 3 lim (\frac{x - 2}{x ^{2} - 9})

\int \frac{1}{x ^{2} + 4 x + 5} d x

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