What is the derivative of (5n^3-3n+1)/(2n-n^4) ?

The derivative of (5n^3-3n+1)/(2n-n^4) is (5n^6-9n^4+24n^3-2)/((2n-n^4)^2)

What is the first derivative of (5n^3-3n+1)/(2n-n^4) ?

The first derivative of (5n^3-3n+1)/(2n-n^4) is (5n^6-9n^4+24n^3-2)/((2n-n^4)^2)

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derivative of

\frac{5 n ^{3} - 3 n + 1}{2 n - n ^{4}}

\frac{5 n ^{6} - 9 n ^{4} + 24 n ^{3} - 2}{( 2 n - n ^{4} ) ^{2}}

Solution steps

\frac{d}{d n} (\frac{5 n ^{3} - 3 n + 1}{2 n - n ^{4}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d n} ( 5 n ^{3} - 3 n + 1 ) ( 2 n - n ^{4} ) - \frac{d}{d n} ( 2 n - n ^{4} ) ( 5 n ^{3} - 3 n + 1 )}{( 2 n - n ^{4} ) ^{2}}

\frac{d}{d n} (5 n^{3} - 3 n + 1) = 15 n^{2} - 3

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

= \frac{( 15 n ^{2} - 3 ) ( 2 n - n ^{4} ) - ( 2 - 4 n ^{3} ) ( 5 n ^{3} - 3 n + 1 )}{( 2 n - n ^{4} ) ^{2}}

Expand

(15 n^{2} - 3) (2 n - n^{4}) - (2 - 4 n^{3}) (5 n^{3} - 3 n + 1) : 5 n^{6} - 9 n^{4} + 24 n^{3} - 2

= \frac{5 n ^{6} - 9 n ^{4} + 24 n ^{3} - 2}{( 2 n - n ^{4} ) ^{2}}

- x^{2} + 3 x + a

x \to 0 lim (2 x^{2} - 2 x - 12)

\int_{- 1}^{1} x^{2} + y^{2} d x

\int sec^{2} (x) tan^{6} (x) d x

\frac{\partial}{\partial y} (x^{2} e^{5 y})

What is the derivative of (5n^3-3n+1)/(2n-n^4) ?
The derivative of (5n^3-3n+1)/(2n-n^4) is (5n^6-9n^4+24n^3-2)/((2n-n^4)^2)
What is the first derivative of (5n^3-3n+1)/(2n-n^4) ?
The first derivative of (5n^3-3n+1)/(2n-n^4) is (5n^6-9n^4+24n^3-2)/((2n-n^4)^2)