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

The derivative of (x^5)/(2-x^4) is (10x^4-x^8)/((2-x^4)^2)

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

The first derivative of (x^5)/(2-x^4) is (10x^4-x^8)/((2-x^4)^2)

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derivative of

\frac{x ^{5}}{2 - x ^{4}}

\frac{10 x ^{4} - x ^{8}}{( 2 - x ^{4} ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{5 x ^{4} ( 2 - x ^{4} ) - ( - 4 x ^{3} ) x ^{5}}{( 2 - x ^{4} ) ^{2}} : \frac{10 x ^{4} - x ^{8}}{( 2 - x ^{4} ) ^{2}}

= \frac{10 x ^{4} - x ^{8}}{( 2 - x ^{4} ) ^{2}}

y = (x^{4} + 1)^{5}

\frac{d x}{d t} = 6 (x^{2} + 1), x (\frac{π}{4}) = 1

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

\frac{d}{d x} (4 ln ((\frac{x}{4})^{2} - 1))

x \to 0 + lim (2 + x^{9} ln (x))

What is the derivative of (x^5)/(2-x^4) ?
The derivative of (x^5)/(2-x^4) is (10x^4-x^8)/((2-x^4)^2)
What is the first derivative of (x^5)/(2-x^4) ?
The first derivative of (x^5)/(2-x^4) is (10x^4-x^8)/((2-x^4)^2)