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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = x^{2} + 4, (- 5, 29)

\frac{d}{d x} (x 2 x - 3)

y = x^{2} - 8 x + 5

x \to 5 lim (1)

\frac{\partial}{\partial x} (ln (4 x + 3 y))

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