What is the derivative of y=(-9e^{4x})/(8x+3) ?

The derivative of y=(-9e^{4x})/(8x+3) is -(9(32e^{4x}x+4e^{4x}))/((8x+3)^2)

What is the first derivative of y=(-9e^{4x})/(8x+3) ?

The first derivative of y=(-9e^{4x})/(8x+3) is -(9(32e^{4x}x+4e^{4x}))/((8x+3)^2)

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derivative of

y = \frac{- 9 e ^{4 x}}{8 x + 3}

- \frac{9 ( 32 e ^{4 x} x + 4 e ^{4 x} )}{( 8 x + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{- 9 e ^{4 x}}{8 x + 3})

Simplify

\frac{- 9 e ^{4 x}}{8 x + 3} : - \frac{9 e ^{4 x}}{8 x + 3}

= \frac{d}{d x} (- \frac{9 e ^{4 x}}{8 x + 3})

Take the constant out:

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

= - 9 \frac{d}{d x} (\frac{e ^{4 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 ^{4 x} ) ( 8 x + 3 ) - \frac{d}{d x} ( 8 x + 3 ) e ^{4 x}}{( 8 x + 3 ) ^{2}}

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

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

= - 9 \cdot \frac{e ^{4 x} \cdot 4 ( 8 x + 3 ) - 8 e ^{4 x}}{( 8 x + 3 ) ^{2}}

Simplify

- 9 \cdot \frac{e ^{4 x} \cdot 4 ( 8 x + 3 ) - 8 e ^{4 x}}{( 8 x + 3 ) ^{2}} : - \frac{9 ( 32 e ^{4 x} x + 4 e ^{4 x} )}{( 8 x + 3 ) ^{2}}

= - \frac{9 ( 32 e ^{4 x} x + 4 e ^{4 x} )}{( 8 x + 3 ) ^{2}}

\int \frac{( x - 2 )}{x - 1} d x

n = 0 \sum \infty \frac{1}{n ^{2} + n}

\int \frac{x}{x ^{2} - 6 x} d x

y^{^{'}} + y = \frac{1}{1 + e ^{t}}

y^{^{'}} + 7 x^{6} y = x^{6}

What is the derivative of y=(-9e^{4x})/(8x+3) ?
The derivative of y=(-9e^{4x})/(8x+3) is -(9(32e^{4x}x+4e^{4x}))/((8x+3)^2)
What is the first derivative of y=(-9e^{4x})/(8x+3) ?
The first derivative of y=(-9e^{4x})/(8x+3) is -(9(32e^{4x}x+4e^{4x}))/((8x+3)^2)