What is the derivative of y= 1/4 x^2-1/2 ln(x) ?

The derivative of y= 1/4 x^2-1/2 ln(x) is x/2-1/(2x)

What is the first derivative of y= 1/4 x^2-1/2 ln(x) ?

The first derivative of y= 1/4 x^2-1/2 ln(x) is x/2-1/(2x)

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derivative of

y = \frac{1}{4} x^{2} - \frac{1}{2} ln (x)

\frac{x}{2} - \frac{1}{2 x}

Solution steps

\frac{d}{d x} (\frac{1}{4} x^{2} - \frac{1}{2} ln (x))

Simplify

\frac{1}{4} x^{2} - \frac{1}{2} ln (x) : \frac{x ^{2}}{4} - \frac{1}{2} ln (x)

= \frac{d}{d x} (\frac{x ^{2}}{4} - \frac{1}{2} ln (x))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{2}}{4}) - \frac{d}{d x} (\frac{1}{2} ln (x))

\frac{d}{d x} (\frac{x ^{2}}{4}) = \frac{x}{2}

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

= \frac{x}{2} - \frac{1}{2 x}

n = 1 \sum \infty \frac{1}{e ^{n^{2}}}

\frac{\partial}{\partial y} (2 x e^{y})

15 - x^{2}

\frac{d}{d t} (\frac{1 - 3 t ^{2}}{1 - 2 t})

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

What is the derivative of y= 1/4 x^2-1/2 ln(x) ?
The derivative of y= 1/4 x^2-1/2 ln(x) is x/2-1/(2x)
What is the first derivative of y= 1/4 x^2-1/2 ln(x) ?
The first derivative of y= 1/4 x^2-1/2 ln(x) is x/2-1/(2x)