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

The derivative of (x^4+2x-1/(5x^2-2x+1)) is (10x^5-6x^4+4x^3-10x^2+10x)/((5x^2-2x+1)^2)

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

The first derivative of (x^4+2x-1/(5x^2-2x+1)) is (10x^5-6x^4+4x^3-10x^2+10x)/((5x^2-2x+1)^2)

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

\frac{10 x ^{5} - 6 x ^{4} + 4 x ^{3} - 10 x ^{2} + 10 x}{( 5 x ^{2} - 2 x + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (5 x^{2} - 2 x + 1) = 10 x - 2

= \frac{( 4 x ^{3} + 2 ) ( 5 x ^{2} - 2 x + 1 ) - ( 10 x - 2 ) ( x ^{4} + 2 x - 1 )}{( 5 x ^{2} - 2 x + 1 ) ^{2}}

Expand

(4 x^{3} + 2) (5 x^{2} - 2 x + 1) - (10 x - 2) (x^{4} + 2 x - 1) : 10 x^{5} - 6 x^{4} + 4 x^{3} - 10 x^{2} + 10 x

= \frac{10 x ^{5} - 6 x ^{4} + 4 x ^{3} - 10 x ^{2} + 10 x}{( 5 x ^{2} - 2 x + 1 ) ^{2}}

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

\frac{\partial}{\partial x} (x y^{3} - x)

\int \frac{1}{324 - x ^{2}} d x

\int tan^{8} (7 x) sec^{4} (7 x) d x

f (x) = x^{2} - 1, at x = - 3

What is the derivative of (x^4+2x-1/(5x^2-2x+1)) ?
The derivative of (x^4+2x-1/(5x^2-2x+1)) is (10x^5-6x^4+4x^3-10x^2+10x)/((5x^2-2x+1)^2)
What is the first derivative of (x^4+2x-1/(5x^2-2x+1)) ?
The first derivative of (x^4+2x-1/(5x^2-2x+1)) is (10x^5-6x^4+4x^3-10x^2+10x)/((5x^2-2x+1)^2)