What is the ((x+1)^{1/2})^' ?

The ((x+1)^{1/2})^' is 1/(2(x+1)^{1/2)}

What is the first ((x+1)^{1/2})^' ?

The first ((x+1)^{1/2})^' is 1/(2(x+1)^{1/2)}

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((x + 1)^{\frac{1}{2}})^{^{'}}

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

Solution steps

((x + 1)^{\frac{1}{2}})^{'}

Apply the chain rule:

\frac{1}{2 ( x + 1 ) ^{\frac{1}{2}}} (x + 1)^{'}

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

(x + 1)^{'} = 1

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

Simplify

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

(x - 7)^{2} (x - 8)^{2}

(7.57 \times 1 0^{2}) (1.1 \times 1 0^{5})

63 x 1 0^{6} (\frac{700000}{1.471} x 1 0^{8})

4 b^{2} (2 b^{5})

459000 e^{- 0.016 (2)}

What is the ((x+1)^{1/2})^' ?
The ((x+1)^{1/2})^' is 1/(2(x+1)^{1/2)}
What is the first ((x+1)^{1/2})^' ?
The first ((x+1)^{1/2})^' is 1/(2(x+1)^{1/2)}