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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

\frac{d u}{d u} = 1

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

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

Simplify

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

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

f (x) = x + 2, at x = 2

x^{3} + x^{2} + \frac{6}{x}

\int \frac{8 x}{x ^{2} - 3} d x

\int \frac{1}{sin ^{2} ( 2 x ) cos ^{2} ( 2 x )} d x

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