What is the derivative of f(x)=((4-2x)^5)/(x^3+12) ?

The derivative of f(x)=((4-2x)^5)/(x^3+12) is (-10(4-2x)^4(x^3+12)-3x^2(4-2x)^5)/((x^3+12)^2)

What is the first derivative of f(x)=((4-2x)^5)/(x^3+12) ?

The first derivative of f(x)=((4-2x)^5)/(x^3+12) is (-10(4-2x)^4(x^3+12)-3x^2(4-2x)^5)/((x^3+12)^2)

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derivative of

f (x) = \frac{( 4 - 2 x ) ^{5}}{x ^{3} + 12}

\frac{- 10 ( 4 - 2 x ) ^{4} ( x ^{3} + 12 ) - 3 x ^{2} ( 4 - 2 x ) ^{5}}{( x ^{3} + 12 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( 4 - 2 x ) ^{5}}{x ^{3} + 12})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ( 4 - 2 x ) ^{5} ) ( x ^{3} + 12 ) - \frac{d}{d x} ( x ^{3} + 12 ) ( 4 - 2 x ) ^{5}}{( x ^{3} + 12 ) ^{2}}

\frac{d}{d x} ((4 - 2 x)^{5}) = - 10 (4 - 2 x)^{4}

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

= \frac{( - 10 ( 4 - 2 x ) ^{4} ) ( x ^{3} + 12 ) - 3 x ^{2} ( 4 - 2 x ) ^{5}}{( x ^{3} + 12 ) ^{2}}

Simplify

= \frac{- 10 ( 4 - 2 x ) ^{4} ( x ^{3} + 12 ) - 3 x ^{2} ( 4 - 2 x ) ^{5}}{( x ^{3} + 12 ) ^{2}}

x \to \infty lim (x^{2} - 3)

y^{^{''}} - 2 y + 6 = 0

\frac{\partial}{\partial x} (e^{- 6 y} sin (6 x))

\frac{d}{d x} (x^{6} ln (x))

\frac{d}{d x} (- 2 cos (x) sin (x) - 2 cos (x))

What is the derivative of f(x)=((4-2x)^5)/(x^3+12) ?
The derivative of f(x)=((4-2x)^5)/(x^3+12) is (-10(4-2x)^4(x^3+12)-3x^2(4-2x)^5)/((x^3+12)^2)
What is the first derivative of f(x)=((4-2x)^5)/(x^3+12) ?
The first derivative of f(x)=((4-2x)^5)/(x^3+12) is (-10(4-2x)^4(x^3+12)-3x^2(4-2x)^5)/((x^3+12)^2)