What is the derivative of (x^3-3xln(2x+1)) ?

The derivative of (x^3-3xln(2x+1)) is ln(2x+1)(3x^2-3)+(2(x^3-3x))/(2x+1)

What is the first derivative of (x^3-3xln(2x+1)) ?

The first derivative of (x^3-3xln(2x+1)) is ln(2x+1)(3x^2-3)+(2(x^3-3x))/(2x+1)

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3} - 3 x, g = ln (2 x + 1)

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

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

y^{^{''}} - 2 y^{^{'}} + 10 y = x^{2} \cdot e^{x}

t an g e n t f (x) = (5 x - x^{2}) (5 - x - x^{2}), at x = 1

t an g e n t f (x) = x - 3

l a pl a ce t r an s f or m \frac{1}{2} cos (2 t)

\int \frac{1}{u a ^{2} - u ^{2}} d u

What is the derivative of (x^3-3xln(2x+1)) ?
The derivative of (x^3-3xln(2x+1)) is ln(2x+1)(3x^2-3)+(2(x^3-3x))/(2x+1)
What is the first derivative of (x^3-3xln(2x+1)) ?
The first derivative of (x^3-3xln(2x+1)) is ln(2x+1)(3x^2-3)+(2(x^3-3x))/(2x+1)