What is the derivative of (6x^2+1/(x^2+9)) ?

The derivative of (6x^2+1/(x^2+9)) is (106x)/((x^2+9)^2)

What is the first derivative of (6x^2+1/(x^2+9)) ?

The first derivative of (6x^2+1/(x^2+9)) is (106x)/((x^2+9)^2)

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\frac{d}{d x} (\frac{6 x ^{2} + 1}{x ^{2} + 9})

\frac{106 x}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x ^{2} + 1}{x ^{2} + 9})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 x ^{2} + 1 ) ( x ^{2} + 9 ) - \frac{d}{d x} ( x ^{2} + 9 ) ( 6 x ^{2} + 1 )}{( x ^{2} + 9 ) ^{2}}

\frac{d}{d x} (6 x^{2} + 1) = 12 x

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

= \frac{12 x ( x ^{2} + 9 ) - 2 x ( 6 x ^{2} + 1 )}{( x ^{2} + 9 ) ^{2}}

Expand

12 x (x^{2} + 9) - 2 x (6 x^{2} + 1) : 106 x

= \frac{106 x}{( x ^{2} + 9 ) ^{2}}

\int \frac{y}{( y + 7 ) ( 2 y - 1 )} d y

- y^{^{''}} = 1

(t^{2} + 16) y^{^{'}} + 2 t y = t^{2} (t^{2} + 16)

\int \frac{5 x ^{3} - 9 x}{x} d x

\int \frac{25}{( x - 1 ) ( x ^{2} + 4 )} d x

What is the derivative of (6x^2+1/(x^2+9)) ?
The derivative of (6x^2+1/(x^2+9)) is (106x)/((x^2+9)^2)
What is the first derivative of (6x^2+1/(x^2+9)) ?
The first derivative of (6x^2+1/(x^2+9)) is (106x)/((x^2+9)^2)