What is the derivative of (x^2-9/(x^2+9)) ?

The derivative of (x^2-9/(x^2+9)) is (36x)/((x^2+9)^2)

What is the first derivative of (x^2-9/(x^2+9)) ?

The first derivative of (x^2-9/(x^2+9)) is (36x)/((x^2+9)^2)

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\frac{d}{d x} (\frac{x ^{2} - 9}{x ^{2} + 9})

\frac{36 x}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 9}{x ^{2} + 9})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 9 ) ( x ^{2} + 9 ) - \frac{d}{d x} ( x ^{2} + 9 ) ( x ^{2} - 9 )}{( x ^{2} + 9 ) ^{2}}

\frac{d}{d x} (x^{2} - 9) = 2 x

\frac{d}{d x} (x^{2} + 9) = 2 x

= \frac{2 x ( x ^{2} + 9 ) - 2 x ( x ^{2} - 9 )}{( x ^{2} + 9 ) ^{2}}

Expand

2 x (x^{2} + 9) - 2 x (x^{2} - 9) : 36 x

= \frac{36 x}{( x ^{2} + 9 ) ^{2}}

x^{^{''}} + x^{^{'}} + 2 x = 0

\frac{d y}{d x} = 2 y (3 - y)

y = 5 arctan (x - 1 + x^{2})

\int \frac{1}{y ^{2} - 3 y} d y

y = - x + 4, y = \frac{x ^{2}}{3}

What is the derivative of (x^2-9/(x^2+9)) ?
The derivative of (x^2-9/(x^2+9)) is (36x)/((x^2+9)^2)
What is the first derivative of (x^2-9/(x^2+9)) ?
The first derivative of (x^2-9/(x^2+9)) is (36x)/((x^2+9)^2)