What is the derivative of f(x)=(x^2-6x+1)/(x^2+4) ?

The derivative of f(x)=(x^2-6x+1)/(x^2+4) is (6x^2+6x-24)/((x^2+4)^2)

What is the first derivative of f(x)=(x^2-6x+1)/(x^2+4) ?

The first derivative of f(x)=(x^2-6x+1)/(x^2+4) is (6x^2+6x-24)/((x^2+4)^2)

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derivative of

f (x) = \frac{x ^{2} - 6 x + 1}{x ^{2} + 4}

\frac{6 x ^{2} + 6 x - 24}{( x ^{2} + 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 6 x + 1}{x ^{2} + 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 6 x + 1 ) ( x ^{2} + 4 ) - \frac{d}{d x} ( x ^{2} + 4 ) ( x ^{2} - 6 x + 1 )}{( x ^{2} + 4 ) ^{2}}

\frac{d}{d x} (x^{2} - 6 x + 1) = 2 x - 6

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

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

Expand

(2 x - 6) (x^{2} + 4) - 2 x (x^{2} - 6 x + 1) : 6 x^{2} + 6 x - 24

= \frac{6 x ^{2} + 6 x - 24}{( x ^{2} + 4 ) ^{2}}

\int (x + 3)^{2} (2 - x)^{5} d x

y^{^{''}} + 6 y + 9 = 0, y (0) = - 3, y^{^{'}} (0) = - 1

\int 4 x^{4} d x

\frac{\partial}{\partial z} (x e^{y} + y e^{z} - z e^{x})

\frac{d}{d x} (8 sin^{6} (x))

What is the derivative of f(x)=(x^2-6x+1)/(x^2+4) ?
The derivative of f(x)=(x^2-6x+1)/(x^2+4) is (6x^2+6x-24)/((x^2+4)^2)
What is the first derivative of f(x)=(x^2-6x+1)/(x^2+4) ?
The first derivative of f(x)=(x^2-6x+1)/(x^2+4) is (6x^2+6x-24)/((x^2+4)^2)