What is the derivative of (x^2+4x-5/(ln(x))) ?

The derivative of (x^2+4x-5/(ln(x))) is (2x^2ln(x)+4xln(x)-x^2-4x+5)/(xln^2(x))

What is the first derivative of (x^2+4x-5/(ln(x))) ?

The first derivative of (x^2+4x-5/(ln(x))) is (2x^2ln(x)+4xln(x)-x^2-4x+5)/(xln^2(x))

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\frac{d}{d x} (\frac{x ^{2} + 4 x - 5}{ln ( x )})

\frac{2 x ^{2} ln ( x ) + 4 x ln ( x ) - x ^{2} - 4 x + 5}{x ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 4 x - 5}{ln ( x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 4 x - 5 ) ln ( x ) - \frac{d}{d x} ( ln ( x ) ) ( x ^{2} + 4 x - 5 )}{( ln ( x ) ) ^{2}}

\frac{d}{d x} (x^{2} + 4 x - 5) = 2 x + 4

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

= \frac{( 2 x + 4 ) ln ( x ) - \frac{1}{x} ( x ^{2} + 4 x - 5 )}{( ln ( x ) ) ^{2}}

Simplify

\frac{( 2 x + 4 ) ln ( x ) - \frac{1}{x} ( x ^{2} + 4 x - 5 )}{ln ^{2} ( x )} : \frac{2 x ^{2} ln ( x ) + 4 x ln ( x ) - x ^{2} - 4 x + 5}{x ln ^{2} ( x )}

= \frac{2 x ^{2} ln ( x ) + 4 x ln ( x ) - x ^{2} - 4 x + 5}{x ln ^{2} ( x )}

n = 1 \sum \infty n! e^{- n}

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

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

\frac{\partial}{\partial x} (\frac{5 x y}{x - y})

f (x) = \frac{6 x}{( 5 - 2 x ) ^{4}}, at x = 1

What is the derivative of (x^2+4x-5/(ln(x))) ?
The derivative of (x^2+4x-5/(ln(x))) is (2x^2ln(x)+4xln(x)-x^2-4x+5)/(xln^2(x))
What is the first derivative of (x^2+4x-5/(ln(x))) ?
The first derivative of (x^2+4x-5/(ln(x))) is (2x^2ln(x)+4xln(x)-x^2-4x+5)/(xln^2(x))