What is the derivative of 8/(sqrt(x^2+6x)) ?

The derivative of 8/(sqrt(x^2+6x)) is -(4(2x+6))/((x^2+6x)^{3/2)}

What is the first derivative of 8/(sqrt(x^2+6x)) ?

The first derivative of 8/(sqrt(x^2+6x)) is -(4(2x+6))/((x^2+6x)^{3/2)}

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\frac{d}{d x} (\frac{8}{x ^{2} + 6 x})

- \frac{4 ( 2 x + 6 )}{( x ^{2} + 6 x ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (\frac{8}{x ^{2} + 6 x})

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{1}{x ^{2} + 6 x})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 8 \frac{d}{d x} ((x^{2} + 6 x)^{- \frac{1}{2}})

Apply the chain rule:

- \frac{1}{2 ( x ^{2} + 6 x ) ^{\frac{3}{2}}} \frac{d}{d x} (x^{2} + 6 x)

= - \frac{1}{2 ( x ^{2} + 6 x ) ^{\frac{3}{2}}} \frac{d}{d x} (x^{2} + 6 x)

\frac{d}{d x} (x^{2} + 6 x) = 2 x + 6

= 8 (- \frac{1}{2 ( x ^{2} + 6 x ) ^{\frac{3}{2}}} (2 x + 6))

Simplify

8 (- \frac{1}{2 ( x ^{2} + 6 x ) ^{\frac{3}{2}}} (2 x + 6)) : - \frac{4 ( 2 x + 6 )}{( x ^{2} + 6 x ) ^{\frac{3}{2}}}

= - \frac{4 ( 2 x + 6 )}{( x ^{2} + 6 x ) ^{\frac{3}{2}}}

x \to 0 lim (3 e^{3 x})

\int_{0}^{1} f (x) d x

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

\int_{- 3}^{5} (4) d x

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

What is the derivative of 8/(sqrt(x^2+6x)) ?
The derivative of 8/(sqrt(x^2+6x)) is -(4(2x+6))/((x^2+6x)^{3/2)}
What is the first derivative of 8/(sqrt(x^2+6x)) ?
The first derivative of 8/(sqrt(x^2+6x)) is -(4(2x+6))/((x^2+6x)^{3/2)}