What is the d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) ?

The d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) is (8{z}+30{x})/((8{z)-5{y})^2}

What is the first d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) ?

The first d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) is (8{z}+30{x})/((8{z)-5{y})^2}

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\frac{d}{d y} (\frac{6 x + y}{8 z - 5 y})

\frac{8 z + 30 x}{( 8 z - 5 y ) ^{2}}

Solution steps

\frac{d}{d y} (\frac{6 x + y}{8 z - 5 y})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d y} ( 6 x + y ) ( 8 z - 5 y ) - \frac{d}{d y} ( 8 z - 5 y ) ( 6 x + y )}{( 8 z - 5 y ) ^{2}}

\frac{d}{d y} (6 x + y) = 1

\frac{d}{d y} (8 z - 5 y) = - 5

= \frac{1 \cdot ( 8 z - 5 y ) - ( - 5 ) ( 6 x + y )}{( 8 z - 5 y ) ^{2}}

Simplify

\frac{1 \cdot ( 8 z - 5 y ) - ( - 5 ) ( 6 x + y )}{( 8 z - 5 y ) ^{2}} : \frac{8 z + 30 x}{( 8 z - 5 y ) ^{2}}

= \frac{8 z + 30 x}{( 8 z - 5 y ) ^{2}}

\frac{d}{d x} (- 4 ln (x))

\frac{\partial}{\partial x} (4 + cos (0.4 x - t))

(e^{x})^{3}

(4 x^{3} + 3 x^{2} + 3 y - 1) d x + (3 x + 2 y + 1) d y = 0

x \to \infty lim (\frac{5 e ^{- 3 x} + 2}{e ^{- 5 x} + 4})

What is the d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) ?
The d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) is (8{z}+30{x})/((8{z)-5{y})^2}
What is the first d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) ?
The first d/(d{y)}((6{x}+{y})/(8{z)-5{y}}) is (8{z}+30{x})/((8{z)-5{y})^2}