What is the derivative of (-3x+8/(2sqrt(4-x))) ?

The derivative of (-3x+8/(2sqrt(4-x))) is (3x-16)/(4(4-x)sqrt(4-x))

What is the first derivative of (-3x+8/(2sqrt(4-x))) ?

The first derivative of (-3x+8/(2sqrt(4-x))) is (3x-16)/(4(4-x)sqrt(4-x))

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Popular >

\frac{d}{d x} (\frac{- 3 x + 8}{2 4 - x})

\frac{3 x - 16}{4 ( 4 - x ) 4 - x}

Solution steps

\frac{d}{d x} (\frac{- 3 x + 8}{2 4 - x})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{- 3 x + 8}{4 - x})

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

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

\frac{d}{d x} (4 - x) = - \frac{1}{2 4 - x}

= \frac{1}{2} \cdot \frac{( - 3 ) 4 - x - ( - \frac{1}{2 4 - x} ) ( - 3 x + 8 )}{( 4 - x ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{( - 3 ) 4 - x - ( - \frac{1}{2 4 - x} ) ( - 3 x + 8 )}{( 4 - x ) ^{2}} : \frac{3 x - 16}{4 ( 4 - x ) 4 - x}

= \frac{3 x - 16}{4 ( 4 - x ) 4 - x}

\frac{d}{d x} (\frac{x - 5}{4 x ^{2} e ^{x}})

x \to 0 lim (x^{4} cos (\frac{1}{x}))

(e^{- 2 t})^{^{'}}

\int \frac{- 6 x ^{4} + 6}{x ^{2} + 1} d x

\int \frac{1}{e ^{10 x}} d x

What is the derivative of (-3x+8/(2sqrt(4-x))) ?
The derivative of (-3x+8/(2sqrt(4-x))) is (3x-16)/(4(4-x)sqrt(4-x))
What is the first derivative of (-3x+8/(2sqrt(4-x))) ?
The first derivative of (-3x+8/(2sqrt(4-x))) is (3x-16)/(4(4-x)sqrt(4-x))