What is the derivative of (5x)/(4x^2+1) ?

The derivative of (5x)/(4x^2+1) is (5(-4x^2+1))/((4x^2+1)^2)

What is the first derivative of (5x)/(4x^2+1) ?

The first derivative of (5x)/(4x^2+1) is (5(-4x^2+1))/((4x^2+1)^2)

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derivative of

\frac{5 x}{4 x ^{2} + 1}

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

Solution steps

\frac{d}{d x} (\frac{5 x}{4 x ^{2} + 1})

Take the constant out:

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

= 5 \frac{d}{d x} (\frac{x}{4 x ^{2} + 1})

Apply the Quotient Rule:

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

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

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

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

= 5 \cdot \frac{1 \cdot ( 4 x ^{2} + 1 ) - 8 xx}{( 4 x ^{2} + 1 ) ^{2}}

Simplify

5 \cdot \frac{1 \cdot ( 4 x ^{2} + 1 ) - 8 xx}{( 4 x ^{2} + 1 ) ^{2}} : \frac{5 ( - 4 x ^{2} + 1 )}{( 4 x ^{2} + 1 ) ^{2}}

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

\int \frac{x}{x ^{2} + 2 x + 2} d x

Y (x) = x^{3} - x^{2} - 6 x, Y (x) = 0

f (x) = 7 x

\int e^{- 0.8 x} d x

\frac{\partial}{\partial x} (x^{2} \cdot ln (y))

What is the derivative of (5x)/(4x^2+1) ?
The derivative of (5x)/(4x^2+1) is (5(-4x^2+1))/((4x^2+1)^2)
What is the first derivative of (5x)/(4x^2+1) ?
The first derivative of (5x)/(4x^2+1) is (5(-4x^2+1))/((4x^2+1)^2)