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

The derivative of y=(x^5)/(4-x^4) is (20x^4-x^8)/((4-x^4)^2)

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

The first derivative of y=(x^5)/(4-x^4) is (20x^4-x^8)/((4-x^4)^2)

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derivative of

y = \frac{x ^{5}}{4 - x ^{4}}

\frac{20 x ^{4} - x ^{8}}{( 4 - x ^{4} ) ^{2}}

Solution steps

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

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

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

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

Simplify

\frac{5 x ^{4} ( 4 - x ^{4} ) - ( - 4 x ^{3} ) x ^{5}}{( 4 - x ^{4} ) ^{2}} : \frac{20 x ^{4} - x ^{8}}{( 4 - x ^{4} ) ^{2}}

= \frac{20 x ^{4} - x ^{8}}{( 4 - x ^{4} ) ^{2}}

in v erse l a pl a ce \frac{2}{s ( 5 s + 1 )}

d er i v a t i v e e^{- y^{2}}

\int (8) d x

\int_{0}^{π} \int 1 d x d y

x \to \infty lim (x^{- 5})

What is the derivative of y=(x^5)/(4-x^4) ?
The derivative of y=(x^5)/(4-x^4) is (20x^4-x^8)/((4-x^4)^2)
What is the first derivative of y=(x^5)/(4-x^4) ?
The first derivative of y=(x^5)/(4-x^4) is (20x^4-x^8)/((4-x^4)^2)