What is the derivative of y=(x^5)/(1-x^4) ?

The derivative of y=(x^5)/(1-x^4) is (5x^4-x^8)/((1-x^4)^2)

What is the first derivative of y=(x^5)/(1-x^4) ?

The first derivative of y=(x^5)/(1-x^4) is (5x^4-x^8)/((1-x^4)^2)

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derivative of

y = \frac{x ^{5}}{1 - x ^{4}}

\frac{5 x ^{4} - x ^{8}}{( 1 - x ^{4} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{5}}{1 - 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} ) ( 1 - x ^{4} ) - \frac{d}{d x} ( 1 - x ^{4} ) x ^{5}}{( 1 - x ^{4} ) ^{2}}

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

\frac{d}{d x} (1 - x^{4}) = - 4 x^{3}

= \frac{5 x ^{4} ( 1 - x ^{4} ) - ( - 4 x ^{3} ) x ^{5}}{( 1 - x ^{4} ) ^{2}}

Simplify

\frac{5 x ^{4} ( 1 - x ^{4} ) - ( - 4 x ^{3} ) x ^{5}}{( 1 - x ^{4} ) ^{2}} : \frac{5 x ^{4} - x ^{8}}{( 1 - x ^{4} ) ^{2}}

= \frac{5 x ^{4} - x ^{8}}{( 1 - x ^{4} ) ^{2}}

x \to \infty lim (sin (0))

n = 2 \sum \infty \frac{ln ( n )}{7 n}

n = 0 \sum \infty \frac{2 n}{n ^{2} + 1}

\frac{x}{x ^{2} + 4}

x x^{2} + 1

What is the derivative of y=(x^5)/(1-x^4) ?
The derivative of y=(x^5)/(1-x^4) is (5x^4-x^8)/((1-x^4)^2)
What is the first derivative of y=(x^5)/(1-x^4) ?
The first derivative of y=(x^5)/(1-x^4) is (5x^4-x^8)/((1-x^4)^2)