What is the derivative of g(x)= 2/(x^3)-1/(x^2) ?

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

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

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

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derivative of

g (x) = \frac{2}{x ^{3}} - \frac{1}{x ^{2}}

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

\frac{d y}{d x} = x + x y^{2}

f (x) = x^{3 x}

(t + 2 y + 3) + (2 t + 4 y - 1) \frac{d y}{d t} = 0

y^{^{''}} - \frac{1}{x} y^{^{'}} + \frac{1}{x ^{2}} y = \frac{2}{x}

\frac{d}{d t} (5 t^{2})

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