What is the derivative of $x \cdot x^{\frac{1}{2}}$ $x*x^(1/2)$ ?

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What is the derivative of $x \cdot x^{\frac{1}{2}}$ $x*x^(1/2)$ ?

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Answer 1 · 2016-03-18T15:56:40

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

Explanation:

$\frac{d}{d x} (x \cdot x^{\frac{1}{2}}) = \frac{d}{d x} (x^{\frac{1}{2} + 1}) = \frac{d}{d x} (x^{\frac{3}{2}}) = \frac{3}{2} x^{\frac{3}{2} - 1} = \frac{3}{2} x^{\frac{1}{2}}$ $d/dx(x*x^(1/2))=d/dx(x^(1/2+1))=d/dx(x^(3/2))=3/2x^(3/2-1)=3/2x^(1/2)$

Answer 2 · 2016-03-18T16:01:56

$\frac{d}{d x} (x \cdot x^{\frac{1}{2}}) = \frac{3}{2} \cdot \sqrt{x}$ $d/(d x)(x*x^(1/2))=3/2*sqrt x$

Explanation:

$\frac{d}{d x} (x \cdot x^{\frac{1}{2}}) = \frac{d}{d x} (x) \cdot x^{\frac{1}{2}} + \frac{d}{d x} (x^{\frac{1}{2}}) \cdot x$ $d/(d x)(x*x^(1/2))=d/(d x) (x)*x^(1/2)+d/(d x)(x^(1/2))*x$
$\frac{d}{d x} (x \cdot x^{\frac{1}{2}}) = 1 \cdot x^{\frac{1}{2}} + \frac{1}{2} \cdot (x^{\frac{1}{2} - 1}) \cdot x$ $d/(d x)(x*x^(1/2))=1*x^(1/2)+1/2*(x^(1/2-1))*x$
$\frac{d}{d x} (x \cdot x^{\frac{1}{2}}) = x^{\frac{1}{2}} + \frac{1}{2} \cdot x^{\frac{1}{2}}$ $d/(d x)(x*x^(1/2))=x^(1/2)+1/2*x^(1/2)$
$\frac{d}{d x} (x \cdot x^{\frac{1}{2}}) = x^{\frac{1}{2}} (1 + \frac{1}{2})$ $d/(d x)(x*x^(1/2))=x^(1/2)(1+1/2)$
$\frac{d}{d x} (x \cdot x^{\frac{1}{2}}) = \frac{3}{2} \cdot \sqrt{x}$ $d/(d x)(x*x^(1/2))=3/2*sqrt x$

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