What is the derivative of f(x)= 1/2 (x+1)^{-1/2} ?

The derivative of f(x)= 1/2 (x+1)^{-1/2} is -1/(4(x+1)^{3/2)}

What is the first derivative of f(x)= 1/2 (x+1)^{-1/2} ?

The first derivative of f(x)= 1/2 (x+1)^{-1/2} is -1/(4(x+1)^{3/2)}

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

d er i v a t i v e e^{c o s (3 x)}

y^{^{''}} + 4 y^{^{'}} - y = 0

s l o p e (3.5) (- 1.5)

l a pl a ce t r an s f or m t^{n}

(x + 2 y) d y - y d x = 0

What is the derivative of f(x)= 1/2 (x+1)^{-1/2} ?
The derivative of f(x)= 1/2 (x+1)^{-1/2} is -1/(4(x+1)^{3/2)}
What is the first derivative of f(x)= 1/2 (x+1)^{-1/2} ?
The first derivative of f(x)= 1/2 (x+1)^{-1/2} is -1/(4(x+1)^{3/2)}