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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\frac{d}{d x} (\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}}

3 y^{^{''}} - 30 y^{^{'}} + 72 y = 6 sin (4 x)

\int \frac{3 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\frac{d}{d x} (sin (x^{3}) + 3 x y)

f (x) = \frac{x}{x + 3}, at x = 1

f (x) = e^{6 x} cos (9 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)