What is the derivative of (9e^{9x})/(8x+3) ?

The derivative of (9e^{9x})/(8x+3) is (9(72e^{9x}x+19e^{9x}))/((8x+3)^2)

What is the first derivative of (9e^{9x})/(8x+3) ?

The first derivative of (9e^{9x})/(8x+3) is (9(72e^{9x}x+19e^{9x}))/((8x+3)^2)

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derivative of

\frac{9 e ^{9 x}}{8 x + 3}

\frac{9 ( 72 e ^{9 x} x + 19 e ^{9 x} )}{( 8 x + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 e ^{9 x}}{8 x + 3})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{e ^{9 x}}{8 x + 3})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d}{d x} ( e ^{9 x} ) ( 8 x + 3 ) - \frac{d}{d x} ( 8 x + 3 ) e ^{9 x}}{( 8 x + 3 ) ^{2}}

\frac{d}{d x} (e^{9 x}) = e^{9 x} \cdot 9

\frac{d}{d x} (8 x + 3) = 8

= 9 \cdot \frac{e ^{9 x} \cdot 9 ( 8 x + 3 ) - 8 e ^{9 x}}{( 8 x + 3 ) ^{2}}

Simplify

9 \cdot \frac{e ^{9 x} \cdot 9 ( 8 x + 3 ) - 8 e ^{9 x}}{( 8 x + 3 ) ^{2}} : \frac{9 ( 72 e ^{9 x} x + 19 e ^{9 x} )}{( 8 x + 3 ) ^{2}}

= \frac{9 ( 72 e ^{9 x} x + 19 e ^{9 x} )}{( 8 x + 3 ) ^{2}}

\int - x^{2} + 8 x d x

x \to 3 lim (- \frac{1}{x - 3})

\frac{d}{d x} (50 x - 0.5 x^{2})

x \to - 5 - lim (\frac{3 x + 8}{x + 5})

\frac{d}{d x} (\frac{cos ( x )}{sin ( x )})

What is the derivative of (9e^{9x})/(8x+3) ?
The derivative of (9e^{9x})/(8x+3) is (9(72e^{9x}x+19e^{9x}))/((8x+3)^2)
What is the first derivative of (9e^{9x})/(8x+3) ?
The first derivative of (9e^{9x})/(8x+3) is (9(72e^{9x}x+19e^{9x}))/((8x+3)^2)