What is the derivative of (x^2+5/(x^2+6x)) ?

The derivative of (x^2+5/(x^2+6x)) is (6x^2-10x-30)/((x^2+6x)^2)

What is the first derivative of (x^2+5/(x^2+6x)) ?

The first derivative of (x^2+5/(x^2+6x)) is (6x^2-10x-30)/((x^2+6x)^2)

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\frac{d}{d x} (\frac{x ^{2} + 5}{x ^{2} + 6 x})

\frac{6 x ^{2} - 10 x - 30}{( x ^{2} + 6 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 5}{x ^{2} + 6 x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 5 ) ( x ^{2} + 6 x ) - \frac{d}{d x} ( x ^{2} + 6 x ) ( x ^{2} + 5 )}{( x ^{2} + 6 x ) ^{2}}

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

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

= \frac{2 x ( x ^{2} + 6 x ) - ( 2 x + 6 ) ( x ^{2} + 5 )}{( x ^{2} + 6 x ) ^{2}}

Expand

2 x (x^{2} + 6 x) - (2 x + 6) (x^{2} + 5) : 6 x^{2} - 10 x - 30

= \frac{6 x ^{2} - 10 x - 30}{( x ^{2} + 6 x ) ^{2}}

t an g e n t y = x^{2} - 2 x - 3, (2, - 3)

n = 0 \sum \infty c n (x - 8)^{n}

d er i v a t i v e \frac{x}{3 x ^{2} + 1} - \frac{1}{8}

\frac{\partial}{\partial t} (\frac{t}{2})

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

What is the derivative of (x^2+5/(x^2+6x)) ?
The derivative of (x^2+5/(x^2+6x)) is (6x^2-10x-30)/((x^2+6x)^2)
What is the first derivative of (x^2+5/(x^2+6x)) ?
The first derivative of (x^2+5/(x^2+6x)) is (6x^2-10x-30)/((x^2+6x)^2)