What is the derivative of x/((3x-2^4)) ?

The derivative of x/((3x-2^4)) is (-2-9x)/((3x-2)^5)

What is the first derivative of x/((3x-2^4)) ?

The first derivative of x/((3x-2^4)) is (-2-9x)/((3x-2)^5)

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\frac{d}{d x} (\frac{x}{( 3 x - 2 ) ^{4}})

\frac{- 2 - 9 x}{( 3 x - 2 ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{x}{( 3 x - 2 ) ^{4}})

Apply the Quotient Rule:

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

= \frac{\frac{d x}{d x} ( 3 x - 2 ) ^{4} - \frac{d}{d x} ( ( 3 x - 2 ) ^{4} ) x}{( ( 3 x - 2 ) ^{4} ) ^{2}}

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

\frac{d}{d x} ((3 x - 2)^{4}) = 12 (3 x - 2)^{3}

= \frac{1 \cdot ( 3 x - 2 ) ^{4} - 12 ( 3 x - 2 ) ^{3} x}{( ( 3 x - 2 ) ^{4} ) ^{2}}

Simplify

\frac{1 \cdot ( 3 x - 2 ) ^{4} - 12 ( 3 x - 2 ) ^{3} x}{( ( 3 x - 2 ) ^{4} ) ^{2}} : \frac{- 2 - 9 x}{( 3 x - 2 ) ^{5}}

= \frac{- 2 - 9 x}{( 3 x - 2 ) ^{5}}

\int (sec^{2} (x) + 7) d x

f (x) = - 3 x^{4}, at x = - 1

\int_{- 2}^{3} (x^{2} + 4) - (- x^{2} - 1) d x

x \to \infty lim (8^{- x})

p (x) = 2 x^{4} + 3 x^{3} - 6 x^{2} + 9

What is the derivative of x/((3x-2^4)) ?
The derivative of x/((3x-2^4)) is (-2-9x)/((3x-2)^5)
What is the first derivative of x/((3x-2^4)) ?
The first derivative of x/((3x-2^4)) is (-2-9x)/((3x-2)^5)