What is the derivative of (9(x-3)/((x-1)^3)) ?

The derivative of (9(x-3)/((x-1)^3)) is (18(-x+4))/((x-1)^4)

What is the first derivative of (9(x-3)/((x-1)^3)) ?

The first derivative of (9(x-3)/((x-1)^3)) is (18(-x+4))/((x-1)^4)

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\frac{d}{d x} (\frac{9 ( x - 3 )}{( x - 1 ) ^{3}})

\frac{18 ( - x + 4 )}{( x - 1 ) ^{4}}

Solution steps

\frac{d}{d x} (\frac{9 ( x - 3 )}{( x - 1 ) ^{3}})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{x - 3}{( x - 1 ) ^{3}})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d}{d x} ( x - 3 ) ( x - 1 ) ^{3} - \frac{d}{d x} ( ( x - 1 ) ^{3} ) ( x - 3 )}{( ( x - 1 ) ^{3} ) ^{2}}

\frac{d}{d x} (x - 3) = 1

\frac{d}{d x} ((x - 1)^{3}) = 3 (x - 1)^{2}

= 9 \cdot \frac{1 \cdot ( x - 1 ) ^{3} - 3 ( x - 1 ) ^{2} ( x - 3 )}{( ( x - 1 ) ^{3} ) ^{2}}

Simplify

9 \cdot \frac{1 \cdot ( x - 1 ) ^{3} - 3 ( x - 1 ) ^{2} ( x - 3 )}{( ( x - 1 ) ^{3} ) ^{2}} : \frac{18 ( - x + 4 )}{( x - 1 ) ^{4}}

= \frac{18 ( - x + 4 )}{( x - 1 ) ^{4}}

x \to 5 - lim (\frac{1}{x ^{2} - 25})

\frac{\partial}{\partial t} (\frac{3 s + t}{4 s - t})

2 y cos (x) d x + 3 sin (x) d y = 0

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 2 x - y}

y^{^{'}} + 5 x^{4} y = x^{4}

What is the derivative of (9(x-3)/((x-1)^3)) ?
The derivative of (9(x-3)/((x-1)^3)) is (18(-x+4))/((x-1)^4)
What is the first derivative of (9(x-3)/((x-1)^3)) ?
The first derivative of (9(x-3)/((x-1)^3)) is (18(-x+4))/((x-1)^4)