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

The derivative of (x^5/(3-x^4)) is (15x^4-x^8)/((3-x^4)^2)

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

The first derivative of (x^5/(3-x^4)) is (15x^4-x^8)/((3-x^4)^2)

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

\frac{15 x ^{4} - x ^{8}}{( 3 - x ^{4} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{5 x ^{4} ( 3 - x ^{4} ) - ( - 4 x ^{3} ) x ^{5}}{( 3 - x ^{4} ) ^{2}} : \frac{15 x ^{4} - x ^{8}}{( 3 - x ^{4} ) ^{2}}

= \frac{15 x ^{4} - x ^{8}}{( 3 - x ^{4} ) ^{2}}

\frac{d}{d x} (\frac{sin ^{2} ( x )}{4})

\frac{d}{d x} (e^{x} ln (x^{2}))

\frac{d y}{d x} = \frac{x y}{x ^{2} + 1}

f (x) = - 3 x^{2} + 24 x - 36

y^{^{'}} = \frac{2 x}{1 + 2 y}

What is the derivative of (x^5/(3-x^4)) ?
The derivative of (x^5/(3-x^4)) is (15x^4-x^8)/((3-x^4)^2)
What is the first derivative of (x^5/(3-x^4)) ?
The first derivative of (x^5/(3-x^4)) is (15x^4-x^8)/((3-x^4)^2)