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

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

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

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

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derivative of

g (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}}

\int \frac{x ^{2} - x + 12}{x ^{3} + 3 x} d x

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

\int 2 x^{- \frac{1}{3}} d x

y = x arcsin (x) + 1 - x^{2}

\frac{\partial}{\partial x} (9 sin (2 x + y))

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