What is the (x^3+3x^2-2ln(x))^' ?

The (x^3+3x^2-2ln(x))^' is 3x^2+6x-2/x

What is the first (x^3+3x^2-2ln(x))^' ?

The first (x^3+3x^2-2ln(x))^' is 3x^2+6x-2/x

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(x^{3} + 3 x^{2} - 2 ln (x))^{^{'}}

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

Solution steps

(x^{3} + 3 x^{2} - 2 ln (x))^{'}

Apply the Sum/Difference Rule:

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

= (x^{3})^{^{'}} + (3 x^{2})^{^{'}} - (2 ln (x))^{^{'}}

(x^{3})^{'} = 3 x^{2}

(3 x^{2})^{'} = 6 x

(2 ln (x))^{'} = \frac{2}{x}

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

\int_{0}^{1} x (1 + 2 x^{5}) d x

\int_{1}^{2} \frac{- 9 ln ( x )}{x ^{2}} d x

\int \frac{x + 2}{x - 3} d x

\int (\frac{5 x ^{2} + x - 4}{x ^{2}}) d x

y = - 2 t^{3}, (- 1, 2)

What is the (x^3+3x^2-2ln(x))^' ?
The (x^3+3x^2-2ln(x))^' is 3x^2+6x-2/x
What is the first (x^3+3x^2-2ln(x))^' ?
The first (x^3+3x^2-2ln(x))^' is 3x^2+6x-2/x