What is the derivative of y=(x^3-3x)ln(2x) ?

The derivative of y=(x^3-3x)ln(2x) is ln(2x)(3x^2-3)+x^2-3

What is the first derivative of y=(x^3-3x)ln(2x) ?

The first derivative of y=(x^3-3x)ln(2x) is ln(2x)(3x^2-3)+x^2-3

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derivative of

y = (x^{3} - 3 x) ln (2 x)

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

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

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

y^{^{'}} + y = \frac{e ^{x}}{y ^{2}}

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

\frac{d}{d x} (9 + 4 x)

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

x \to - 2 lim ((3 - x)^{2})

What is the derivative of y=(x^3-3x)ln(2x) ?
The derivative of y=(x^3-3x)ln(2x) is ln(2x)(3x^2-3)+x^2-3
What is the first derivative of y=(x^3-3x)ln(2x) ?
The first derivative of y=(x^3-3x)ln(2x) is ln(2x)(3x^2-3)+x^2-3