What is the derivative of (ax^2/(ln(x/4))) ?

The derivative of (ax^2/(ln(x/4))) is (a(2xln(x/4)-x))/(ln^2(x/4))

What is the first derivative of (ax^2/(ln(x/4))) ?

The first derivative of (ax^2/(ln(x/4))) is (a(2xln(x/4)-x))/(ln^2(x/4))

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

y^{^{'}} = \frac{x - y}{x}

e^{- (x + 2)} - sin (x^{2}) + 1

\frac{x ^{2}}{4} + x + 1

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

y = \frac{5 x}{x - 2}, (3, 15)

What is the derivative of (ax^2/(ln(x/4))) ?
The derivative of (ax^2/(ln(x/4))) is (a(2xln(x/4)-x))/(ln^2(x/4))
What is the first derivative of (ax^2/(ln(x/4))) ?
The first derivative of (ax^2/(ln(x/4))) is (a(2xln(x/4)-x))/(ln^2(x/4))