What is the derivative of (e^{2x-1}/((x^2+5x))) ?

The derivative of (e^{2x-1}/((x^2+5x))) is (2e^{2x-1}x^2+8e^{2x-1}x-5e^{2x-1})/((x^2+5x)^2)

What is the first derivative of (e^{2x-1}/((x^2+5x))) ?

The first derivative of (e^{2x-1}/((x^2+5x))) is (2e^{2x-1}x^2+8e^{2x-1}x-5e^{2x-1})/((x^2+5x)^2)

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\frac{d}{d x} (\frac{e ^{2 x - 1}}{( x ^{2} + 5 x )})

\frac{2 e ^{2 x - 1} x ^{2} + 8 e ^{2 x - 1} x - 5 e ^{2 x - 1}}{( x ^{2} + 5 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{2 x - 1}}{( x ^{2} + 5 x )})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( e ^{2 x - 1} ) ( x ^{2} + 5 x ) - \frac{d}{d x} ( x ^{2} + 5 x ) e ^{2 x - 1}}{( x ^{2} + 5 x ) ^{2}}

\frac{d}{d x} (e^{2 x - 1}) = e^{2 x - 1} \cdot 2

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

= \frac{e ^{2 x - 1} \cdot 2 ( x ^{2} + 5 x ) - ( 2 x + 5 ) e ^{2 x - 1}}{( x ^{2} + 5 x ) ^{2}}

Expand

e^{2 x - 1} \cdot 2 (x^{2} + 5 x) - (2 x + 5) e^{2 x - 1} : 2 e^{2 x - 1} x^{2} + 8 e^{2 x - 1} x - 5 e^{2 x - 1}

= \frac{2 e ^{2 x - 1} x ^{2} + 8 e ^{2 x - 1} x - 5 e ^{2 x - 1}}{( x ^{2} + 5 x ) ^{2}}

1 + 4 - x

\frac{1}{4 y}

2 y d x + x d y = 0

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

\int_{0}^{5} \frac{1}{25 + x ^{2}} d x

What is the derivative of (e^{2x-1}/((x^2+5x))) ?
The derivative of (e^{2x-1}/((x^2+5x))) is (2e^{2x-1}x^2+8e^{2x-1}x-5e^{2x-1})/((x^2+5x)^2)
What is the first derivative of (e^{2x-1}/((x^2+5x))) ?
The first derivative of (e^{2x-1}/((x^2+5x))) is (2e^{2x-1}x^2+8e^{2x-1}x-5e^{2x-1})/((x^2+5x)^2)