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

The derivative of f(x)=(4x^3)/((3-2x)^5) is (4x^2(4x+9))/((3-2x)^6)

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

The first derivative of f(x)=(4x^3)/((3-2x)^5) is (4x^2(4x+9))/((3-2x)^6)

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derivative of

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

\frac{4 x ^{2} ( 4 x + 9 )}{( 3 - 2 x ) ^{6}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

4 \cdot \frac{3 x ^{2} ( 3 - 2 x ) ^{5} - ( - 10 ( 3 - 2 x ) ^{4} ) x ^{3}}{( ( 3 - 2 x ) ^{5} ) ^{2}} : \frac{4 x ^{2} ( 4 x + 9 )}{( 3 - 2 x ) ^{6}}

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

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

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

\frac{72000 + 60 x}{x}

f (x) = cos (sin (x))

t^{2} y^{^{'}} - y = 0

What is the derivative of f(x)=(4x^3)/((3-2x)^5) ?
The derivative of f(x)=(4x^3)/((3-2x)^5) is (4x^2(4x+9))/((3-2x)^6)
What is the first derivative of f(x)=(4x^3)/((3-2x)^5) ?
The first derivative of f(x)=(4x^3)/((3-2x)^5) is (4x^2(4x+9))/((3-2x)^6)