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

The derivative of f(x)=(e^x)/(x^2+9) is (e^x(x^2+9)-2xe^x)/((x^2+9)^2)

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

\frac{d}{d t} (t^{3 t})

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

\frac{d}{d x} (\frac{6}{x ^{2} + 3})

x \to - 2 lim (x + 4)

(x^{2} + 1)^{2}

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