What is the derivative of f(x)=(x^2+6)/(9x-2) ?

The derivative of f(x)=(x^2+6)/(9x-2) is (9x^2-4x-54)/((9x-2)^2)

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

The first derivative of f(x)=(x^2+6)/(9x-2) is (9x^2-4x-54)/((9x-2)^2)

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derivative of

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

\frac{9 x ^{2} - 4 x - 54}{( 9 x - 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

2 x (9 x - 2) - 9 (x^{2} + 6) : 9 x^{2} - 4 x - 54

= \frac{9 x ^{2} - 4 x - 54}{( 9 x - 2 ) ^{2}}

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

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

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

\frac{d x}{d t} = 2 x - x^{2}

\frac{\partial}{\partial x} (3 x - (\frac{3}{4} x + \frac{1}{4} y)^{2})

What is the derivative of f(x)=(x^2+6)/(9x-2) ?
The derivative of f(x)=(x^2+6)/(9x-2) is (9x^2-4x-54)/((9x-2)^2)
What is the first derivative of f(x)=(x^2+6)/(9x-2) ?
The first derivative of f(x)=(x^2+6)/(9x-2) is (9x^2-4x-54)/((9x-2)^2)