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

The (2xln(x)+x^2-1)^' is 2x+2(ln(x)+1)

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

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

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(2 x ln (x) + x^{2} - 1)^{^{'}}

2 x + 2 (ln (x) + 1)

Solution steps

(2 x ln (x) + x^{2} - 1)^{'}

Apply the Sum/Difference Rule:

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

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

(2 x ln (x))^{'} = 2 (ln (x) + 1)

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

1^{'} = 0

= 2 (ln (x) + 1) + 2 x - 0

Simplify

= 2 x + 2 (ln (x) + 1)

\frac{\partial}{\partial x} (u (x))

\int (y^{2} e^{x y}) d x

\frac{\partial}{\partial x} (\frac{x - 2 y}{x + 2 y})

\frac{\partial}{\partial x} (ln (x + 2 c t))

What is the (2xln(x)+x^2-1)^' ?
The (2xln(x)+x^2-1)^' is 2x+2(ln(x)+1)
What is the first (2xln(x)+x^2-1)^' ?
The first (2xln(x)+x^2-1)^' is 2x+2(ln(x)+1)