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

The derivative of 2/(sqrt(3-x)) is 1/((3-x)^{3/2)}

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

The first derivative of 2/(sqrt(3-x)) is 1/((3-x)^{3/2)}

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\frac{d}{d x} (\frac{2}{3 - x})

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

Solution steps

\frac{d}{d x} (\frac{2}{3 - x})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int x^{3} x^{4} + 5 d x

\frac{d}{d x} (sec (tan^{2} (x^{4})))

n = 0 \sum \infty (\frac{e}{2 π})^{n}

d y = (x y^{2} + 3 x y) d x

y^{^{''}} + 4 y^{^{'}} + 4 y = 25 cos (x)

What is the derivative of 2/(sqrt(3-x)) ?
The derivative of 2/(sqrt(3-x)) is 1/((3-x)^{3/2)}
What is the first derivative of 2/(sqrt(3-x)) ?
The first derivative of 2/(sqrt(3-x)) is 1/((3-x)^{3/2)}