What is the (1/x+x^2-2)^' ?

The (1/x+x^2-2)^' is 2x-1/(x^2)

What is the first (1/x+x^2-2)^' ?

The first (1/x+x^2-2)^' is 2x-1/(x^2)

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

2 x - \frac{1}{x ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

(x^{2})^{'} = 2 x

2^{'} = 0

= - \frac{1}{x ^{2}} + 2 x - 0

Simplify

= 2 x - \frac{1}{x ^{2}}

\frac{d}{d x} (19 + x)

\frac{\partial}{\partial r} (r - s)

\int e^{3 t} d t

\int y e^{8 y^{2}} d y

\frac{d}{d x} (\frac{x y + 3 x - y - 3}{x y - 2 x + 4 y - 8})