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

The derivative of f(x)=(4x^3)/(2x^2-5) is (4(2x^4-15x^2))/((2x^2-5)^2)

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

The first derivative of f(x)=(4x^3)/(2x^2-5) is (4(2x^4-15x^2))/((2x^2-5)^2)

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derivative of

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

\frac{4 ( 2 x ^{4} - 15 x ^{2} )}{( 2 x ^{2} - 5 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

4 \cdot \frac{3 x ^{2} ( 2 x ^{2} - 5 ) - 4 x x ^{3}}{( 2 x ^{2} - 5 ) ^{2}} : \frac{4 ( 2 x ^{4} - 15 x ^{2} )}{( 2 x ^{2} - 5 ) ^{2}}

= \frac{4 ( 2 x ^{4} - 15 x ^{2} )}{( 2 x ^{2} - 5 ) ^{2}}

4 x + 5 y = 20

f = 4

f (x) = sin (x)

f (x) = 1 + cos (x)

sec^{2} (x)

What is the derivative of f(x)=(4x^3)/(2x^2-5) ?
The derivative of f(x)=(4x^3)/(2x^2-5) is (4(2x^4-15x^2))/((2x^2-5)^2)
What is the first derivative of f(x)=(4x^3)/(2x^2-5) ?
The first derivative of f(x)=(4x^3)/(2x^2-5) is (4(2x^4-15x^2))/((2x^2-5)^2)