What is the derivative of f(x)=(e^x)/(4x^2+1) ?

The derivative of f(x)=(e^x)/(4x^2+1) is (e^x(4x^2+1)-8xe^x)/((4x^2+1)^2)

What is the first derivative of f(x)=(e^x)/(4x^2+1) ?

The first derivative of f(x)=(e^x)/(4x^2+1) is (e^x(4x^2+1)-8xe^x)/((4x^2+1)^2)

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derivative of

f (x) = \frac{e ^{x}}{4 x ^{2} + 1}

\frac{e ^{x} ( 4 x ^{2} + 1 ) - 8 x e ^{x}}{( 4 x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{x}}{4 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} ) ( 4 x ^{2} + 1 ) - \frac{d}{d x} ( 4 x ^{2} + 1 ) e ^{x}}{( 4 x ^{2} + 1 ) ^{2}}

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

\frac{d}{d x} (4 x^{2} + 1) = 8 x

= \frac{e ^{x} ( 4 x ^{2} + 1 ) - 8 x e ^{x}}{( 4 x ^{2} + 1 ) ^{2}}

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

\int \frac{1}{4 + 9 x ^{2}} d x

\frac{d y}{d x} - 4 y = x^{2} + 9

y = 8 x - x^{2} (1.7)

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

What is the derivative of f(x)=(e^x)/(4x^2+1) ?
The derivative of f(x)=(e^x)/(4x^2+1) is (e^x(4x^2+1)-8xe^x)/((4x^2+1)^2)
What is the first derivative of f(x)=(e^x)/(4x^2+1) ?
The first derivative of f(x)=(e^x)/(4x^2+1) is (e^x(4x^2+1)-8xe^x)/((4x^2+1)^2)