What is the derivative of ((k^2x^3-6kx+9))/(x-1) ?

The derivative of ((k^2x^3-6kx+9))/(x-1) is (2k^2x^3-3k^2x^2+6k-9)/((x-1)^2)

What is the first derivative of ((k^2x^3-6kx+9))/(x-1) ?

The first derivative of ((k^2x^3-6kx+9))/(x-1) is (2k^2x^3-3k^2x^2+6k-9)/((x-1)^2)

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derivative of

\frac{( k ^{2} x ^{3} - 6 k x + 9 )}{x - 1}

\frac{2 k ^{2} x ^{3} - 3 k ^{2} x ^{2} + 6 k - 9}{( x - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{k ^{2} x ^{3} - 6 k x + 9}{x - 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( k ^{2} x ^{3} - 6 k x + 9 ) ( x - 1 ) - \frac{d}{d x} ( x - 1 ) ( k ^{2} x ^{3} - 6 k x + 9 )}{( x - 1 ) ^{2}}

\frac{d}{d x} (k^{2} x^{3} - 6 k x + 9) = 3 k^{2} x^{2} - 6 k

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

= \frac{( 3 k ^{2} x ^{2} - 6 k ) ( x - 1 ) - 1 \cdot ( k ^{2} x ^{3} - 6 k x + 9 )}{( x - 1 ) ^{2}}

Simplify

\frac{( 3 k ^{2} x ^{2} - 6 k ) ( x - 1 ) - 1 \cdot ( k ^{2} x ^{3} - 6 k x + 9 )}{( x - 1 ) ^{2}} : \frac{2 k ^{2} x ^{3} - 3 k ^{2} x ^{2} + 6 k - 9}{( x - 1 ) ^{2}}

= \frac{2 k ^{2} x ^{3} - 3 k ^{2} x ^{2} + 6 k - 9}{( x - 1 ) ^{2}}

\int \frac{x}{2 x ^{2} + 12 x + 23} d x

x (x + y) y^{^{'}} = y (x - y)

\int sin (u) cos (u) d u

f (x) = cos (x)

\int \frac{e ^{4 x}}{x} d x

What is the derivative of ((k^2x^3-6kx+9))/(x-1) ?
The derivative of ((k^2x^3-6kx+9))/(x-1) is (2k^2x^3-3k^2x^2+6k-9)/((x-1)^2)
What is the first derivative of ((k^2x^3-6kx+9))/(x-1) ?
The first derivative of ((k^2x^3-6kx+9))/(x-1) is (2k^2x^3-3k^2x^2+6k-9)/((x-1)^2)