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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2 x ln (x)) = 2 (ln (x) + 1)

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

\frac{d}{d x} (1) = 0

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

Simplify

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

\int 3 sin^{5} (x) cos^{- \frac{3}{2}} (x) d x

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

\int \frac{1}{x ( 2 + 3 u )} d x

y^{^{''}} - 2 y^{^{'}} + y = 0

\int x (12 x + 4) d x

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