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

The derivative of y=x^3+ln(2x) is 3x^2+1/x

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

The first derivative of y=x^3+ln(2x) is 3x^2+1/x

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derivative of

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

3 x^{2} + \frac{1}{x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 3 x^{2} + \frac{1}{x}

\int \frac{x tan ( x ^{2} )}{cos ( x ^{2} )} d x

\frac{\partial}{\partial x} (4 y^{3} cos (5 x))

n = 0 \sum \infty \frac{2}{3 ^{n - 1}}

\int - 3 x^{2} d x

\frac{\partial}{\partial z} (\frac{x ^{2}}{y} - x z)

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