What is the d/(du)((5u)/(4u^2+1)) ?

The d/(du)((5u)/(4u^2+1)) is (5(-4u^2+1))/((4u^2+1)^2)

What is the first d/(du)((5u)/(4u^2+1)) ?

The first d/(du)((5u)/(4u^2+1)) is (5(-4u^2+1))/((4u^2+1)^2)

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\frac{d}{d u} (\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}}

y = \frac{1}{x}, at x = 2

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 8 y ^{2}})

\int \frac{1}{ln ( x + 1 )} d x

\int \frac{1}{x ^{2} 10 x + 1} d x

\int e^{θ} sin (θ) d θ

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