What is the derivative of ((17-x)sqrt(2x+8))/2 ?

The derivative of ((17-x)sqrt(2x+8))/2 is (-3x+9)/(2sqrt(2x+8))

What is the first derivative of ((17-x)sqrt(2x+8))/2 ?

The first derivative of ((17-x)sqrt(2x+8))/2 is (-3x+9)/(2sqrt(2x+8))

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derivative of

\frac{( 17 - x ) 2 x + 8}{2}

\frac{- 3 x + 9}{2 2 x + 8}

Solution steps

\frac{d}{d x} (\frac{( 17 - x ) 2 x + 8}{2})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} ((17 - x) 2 x + 8)

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 17 - x, g = 2 x + 8

= \frac{1}{2} (\frac{d}{d x} (17 - x) 2 x + 8 + \frac{d}{d x} (2 x + 8) (17 - x))

\frac{d}{d x} (17 - x) = - 1

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

= \frac{1}{2} ((- 1) 2 x + 8 + \frac{1}{2 x + 8} (17 - x))

Simplify

\frac{1}{2} ((- 1) 2 x + 8 + \frac{1}{2 x + 8} (17 - x)) : \frac{- 3 x + 9}{2 2 x + 8}

= \frac{- 3 x + 9}{2 2 x + 8}

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

\int_{0}^{0.6} (600 x) d x

\frac{d y}{d x} = \frac{( sec ( y ) )}{( x - 1 ) ^{2}}

\int_{0}^{π} sin^{5} (x) d x

What is the derivative of ((17-x)sqrt(2x+8))/2 ?
The derivative of ((17-x)sqrt(2x+8))/2 is (-3x+9)/(2sqrt(2x+8))
What is the first derivative of ((17-x)sqrt(2x+8))/2 ?
The first derivative of ((17-x)sqrt(2x+8))/2 is (-3x+9)/(2sqrt(2x+8))