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

The derivative of ((x^2+1^2)/(4x)) is (3x^4+2x^2-1)/(4x^2)

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

The first derivative of ((x^2+1^2)/(4x)) is (3x^4+2x^2-1)/(4x^2)

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

\frac{3 x ^{4} + 2 x ^{2} - 1}{4 x ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

\frac{d x}{d x} = 1

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

Simplify

\frac{1}{4} \cdot \frac{4 x ( x ^{2} + 1 ) x - 1 \cdot ( x ^{2} + 1 ) ^{2}}{x ^{2}} : \frac{3 x ^{4} + 2 x ^{2} - 1}{4 x ^{2}}

= \frac{3 x ^{4} + 2 x ^{2} - 1}{4 x ^{2}}

\frac{\partial}{\partial x} (\frac{x + y}{1 + x ^{2}})

\int \frac{e ^{4 x}}{1 + e ^{4 x}} d x

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

\frac{\partial}{\partial x} (sin (π (6 x - 6 y)))

What is the derivative of ((x^2+1^2)/(4x)) ?
The derivative of ((x^2+1^2)/(4x)) is (3x^4+2x^2-1)/(4x^2)
What is the first derivative of ((x^2+1^2)/(4x)) ?
The first derivative of ((x^2+1^2)/(4x)) is (3x^4+2x^2-1)/(4x^2)