What is the derivative of 3/(sqrt(6-x)) ?

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

What is the first derivative of 3/(sqrt(6-x)) ?

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

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\frac{d}{d x} (\frac{3}{6 - x})

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

Solution steps

\frac{d}{d x} (\frac{3}{6 - x})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (6 - x) = - 1

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

Simplify

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

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

\frac{d}{d x} (3 x^{4} + 15)

e^{6 x} cos (9 x)

y = \frac{4}{x}, (- 4, - 1)

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

\frac{\partial}{\partial x} (x^{5} + 4 z^{2} e^{y - x})

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