What is the derivative of (pi(9x^2+6x+1)/(36x)) ?

The derivative of (pi(9x^2+6x+1)/(36x)) is (pi(9x^2-1))/(36x^2)

What is the first derivative of (pi(9x^2+6x+1)/(36x)) ?

The first derivative of (pi(9x^2+6x+1)/(36x)) is (pi(9x^2-1))/(36x^2)

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\frac{d}{d x} (\frac{π ( 9 x ^{2} + 6 x + 1 )}{36 x})

\frac{π ( 9 x ^{2} - 1 )}{36 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{π ( 9 x ^{2} + 6 x + 1 )}{36 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{π}{36} \frac{d}{d x} (\frac{9 x ^{2} + 6 x + 1}{x})

Apply the Quotient Rule:

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

= \frac{π}{36} \cdot \frac{\frac{d}{d x} ( 9 x ^{2} + 6 x + 1 ) x - \frac{d x}{d x} ( 9 x ^{2} + 6 x + 1 )}{x ^{2}}

\frac{d}{d x} (9 x^{2} + 6 x + 1) = 18 x + 6

\frac{d x}{d x} = 1

= \frac{π}{36} \cdot \frac{( 18 x + 6 ) x - 1 \cdot ( 9 x ^{2} + 6 x + 1 )}{x ^{2}}

Simplify

\frac{π}{36} \cdot \frac{( 18 x + 6 ) x - 1 \cdot ( 9 x ^{2} + 6 x + 1 )}{x ^{2}} : \frac{π ( 9 x ^{2} - 1 )}{36 x ^{2}}

= \frac{π ( 9 x ^{2} - 1 )}{36 x ^{2}}

\frac{8}{3 x ^{2}}

tan^{9} (x)

x \to 2 - lim (\frac{x + 1}{4 - x ^{2}})

\int 3 - t^{2} d t

f (t) = \frac{2 t + 1}{t + 3}

What is the derivative of (pi(9x^2+6x+1)/(36x)) ?
The derivative of (pi(9x^2+6x+1)/(36x)) is (pi(9x^2-1))/(36x^2)
What is the first derivative of (pi(9x^2+6x+1)/(36x)) ?
The first derivative of (pi(9x^2+6x+1)/(36x)) is (pi(9x^2-1))/(36x^2)