What is the derivative of (5x/(1+2x^2)) ?

The derivative of (5x/(1+2x^2)) is (5(1-2x^2))/((1+2x^2)^2)

What is the first derivative of (5x/(1+2x^2)) ?

The first derivative of (5x/(1+2x^2)) is (5(1-2x^2))/((1+2x^2)^2)

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\frac{d}{d x} (\frac{5 x}{1 + 2 x ^{2}})

\frac{5 ( 1 - 2 x ^{2} )}{( 1 + 2 x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 x}{1 + 2 x ^{2}})

Take the constant out:

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

= 5 \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}}

= 5 \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

= 5 \cdot \frac{1 \cdot ( 1 + 2 x ^{2} ) - 4 xx}{( 1 + 2 x ^{2} ) ^{2}}

Simplify

5 \cdot \frac{1 \cdot ( 1 + 2 x ^{2} ) - 4 xx}{( 1 + 2 x ^{2} ) ^{2}} : \frac{5 ( 1 - 2 x ^{2} )}{( 1 + 2 x ^{2} ) ^{2}}

= \frac{5 ( 1 - 2 x ^{2} )}{( 1 + 2 x ^{2} ) ^{2}}

x \to 0 + lim (\frac{sin ^{2} ( x )}{x})

x^{2} + x y - y^{2} = - 1, (1, 2)

(x - 7) (2 x + 4)

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

y = (4 x^{3} + 5) (x^{4} - 3 x)

What is the derivative of (5x/(1+2x^2)) ?
The derivative of (5x/(1+2x^2)) is (5(1-2x^2))/((1+2x^2)^2)
What is the first derivative of (5x/(1+2x^2)) ?
The first derivative of (5x/(1+2x^2)) is (5(1-2x^2))/((1+2x^2)^2)