What is the d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) ?

The d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) is (8(-3{x}^2+1){y)/z}{({x}^2+1)^3}

What is the first d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) ?

The first d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) is (8(-3{x}^2+1){y)/z}{({x}^2+1)^3}

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\frac{d}{d x} (\frac{8 x y z}{( x ^{2} + 1 ) ^{2}})

\frac{8 ( - 3 x ^{2} + 1 ) y z}{( x ^{2} + 1 ) ^{3}}

Solution steps

\frac{d}{d x} (\frac{8 x y z}{( x ^{2} + 1 ) ^{2}})

Take the constant out:

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

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

Apply the Quotient Rule:

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

= 8 y z \frac{\frac{d}{d x} ( x ) ( x ^{2} + 1 ) ^{2} - \frac{d}{d x} ( ( x ^{2} + 1 ) ^{2} ) x}{( ( x ^{2} + 1 ) ^{2} ) ^{2}}

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

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

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

Simplify

8 y z \frac{1 \cdot ( x ^{2} + 1 ) ^{2} - 4 x ( x ^{2} + 1 ) x}{( ( x ^{2} + 1 ) ^{2} ) ^{2}} : \frac{8 ( - 3 x ^{2} + 1 ) y z}{( x ^{2} + 1 ) ^{3}}

= \frac{8 ( - 3 x ^{2} + 1 ) y z}{( x ^{2} + 1 ) ^{3}}

\int x^{3} - 3 x^{2} - x + 3 d x

\frac{\partial}{\partial y} (3)

\int sin^{2} (6 x) d x

\frac{\partial}{\partial y} (y cos (yz))

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

What is the d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) ?
The d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) is (8(-3{x}^2+1){y)/z}{({x}^2+1)^3}
What is the first d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) ?
The first d/(d{x)}((8{x)/y {z}}{({x}^2+1)^2}) is (8(-3{x}^2+1){y)/z}{({x}^2+1)^3}