What is the derivative of (e^x/(9e^x+x^2)) ?

The derivative of (e^x/(9e^x+x^2)) is (e^xx^2-2e^xx)/((9e^x+x^2)^2)

What is the first derivative of (e^x/(9e^x+x^2)) ?

The first derivative of (e^x/(9e^x+x^2)) is (e^xx^2-2e^xx)/((9e^x+x^2)^2)

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\frac{d}{d x} (\frac{e ^{x}}{9 e ^{x} + x ^{2}})

\frac{e ^{x} x ^{2} - 2 e ^{x} x}{( 9 e ^{x} + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{x}}{9 e ^{x} + x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( e ^{x} ) ( 9 e ^{x} + x ^{2} ) - \frac{d}{d x} ( 9 e ^{x} + x ^{2} ) e ^{x}}{( 9 e ^{x} + x ^{2} ) ^{2}}

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

\frac{d}{d x} (9 e^{x} + x^{2}) = 9 e^{x} + 2 x

= \frac{e ^{x} ( 9 e ^{x} + x ^{2} ) - ( 9 e ^{x} + 2 x ) e ^{x}}{( 9 e ^{x} + x ^{2} ) ^{2}}

Expand

e^{x} (9 e^{x} + x^{2}) - (9 e^{x} + 2 x) e^{x} : e^{x} x^{2} - 2 e^{x} x

= \frac{e ^{x} x ^{2} - 2 e ^{x} x}{( 9 e ^{x} + x ^{2} ) ^{2}}

a re a - x^{2} + 10 x - 16, 2 x - 4, 2, 8

\frac{d}{d x} (2 arcsin (x - 1))

in v erse l a pl a ce \frac{20}{s ( s ^{2} + 6 s + 144 )}

a re a y = \frac{3}{x}, y = \frac{3}{x ^{2}}, x = 4

d er i v a t i v e f (x) = 800 - 800 x^{- 1}

What is the derivative of (e^x/(9e^x+x^2)) ?
The derivative of (e^x/(9e^x+x^2)) is (e^xx^2-2e^xx)/((9e^x+x^2)^2)
What is the first derivative of (e^x/(9e^x+x^2)) ?
The first derivative of (e^x/(9e^x+x^2)) is (e^xx^2-2e^xx)/((9e^x+x^2)^2)