What is the derivative of (3x+36/(2sqrt(x+18))) ?

The derivative of (3x+36/(2sqrt(x+18))) is (3x+72)/(4(x+18)sqrt(x+18))

What is the first derivative of (3x+36/(2sqrt(x+18))) ?

The first derivative of (3x+36/(2sqrt(x+18))) is (3x+72)/(4(x+18)sqrt(x+18))

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\frac{d}{d x} (\frac{3 x + 36}{2 x + 18})

\frac{3 x + 72}{4 ( x + 18 ) x + 18}

Solution steps

\frac{d}{d x} (\frac{3 x + 36}{2 x + 18})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{3 x + 36}{x + 18})

Apply the Quotient Rule:

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

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( 3 x + 36 ) x + 18 - \frac{d}{d x} ( x + 18 ) ( 3 x + 36 )}{( x + 18 ) ^{2}}

\frac{d}{d x} (3 x + 36) = 3

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

= \frac{1}{2} \cdot \frac{3 x + 18 - \frac{1}{2 x + 18} ( 3 x + 36 )}{( x + 18 ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{3 x + 18 - \frac{1}{2 x + 18} ( 3 x + 36 )}{( x + 18 ) ^{2}} : \frac{3 x + 72}{4 ( x + 18 ) x + 18}

= \frac{3 x + 72}{4 ( x + 18 ) x + 18}

(x + y) y^{^{'}} = y

\int \frac{e ^{x}}{( e ^{x} - 7 ) ( e ^{2 x} + 1 )} d x

f (x) = x cos (2 x)

\int x x + 11 d x

\frac{4}{2 + 2 x ^{2}}

What is the derivative of (3x+36/(2sqrt(x+18))) ?
The derivative of (3x+36/(2sqrt(x+18))) is (3x+72)/(4(x+18)sqrt(x+18))
What is the first derivative of (3x+36/(2sqrt(x+18))) ?
The first derivative of (3x+36/(2sqrt(x+18))) is (3x+72)/(4(x+18)sqrt(x+18))