What is the derivative of f(x)=((e^x))/(x^2+1) ?

The derivative of f(x)=((e^x))/(x^2+1) is (e^x(x^2+1)-2xe^x)/((x^2+1)^2)

What is the first derivative of f(x)=((e^x))/(x^2+1) ?

The first derivative of f(x)=((e^x))/(x^2+1) is (e^x(x^2+1)-2xe^x)/((x^2+1)^2)

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derivative of

f (x) = \frac{( e ^{x} )}{x ^{2} + 1}

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

Solution steps

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

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

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

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

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

x \to 3 lim (\frac{5 x}{6 - 2 x})

\int \frac{( x ^{2} )}{9 - x ^{2}} d x

x \to 0 lim (\frac{1}{x ^{3}})

\frac{d}{d x} (\frac{4}{1 + x})

y^{^{''}} - 6 y^{^{'}} + 45 y = 0

What is the derivative of f(x)=((e^x))/(x^2+1) ?
The derivative of f(x)=((e^x))/(x^2+1) is (e^x(x^2+1)-2xe^x)/((x^2+1)^2)
What is the first derivative of f(x)=((e^x))/(x^2+1) ?
The first derivative of f(x)=((e^x))/(x^2+1) is (e^x(x^2+1)-2xe^x)/((x^2+1)^2)