What is the derivative of (xy^{1/2}) ?

The derivative of (xy^{1/2}) is (y+x\frac{dy)/(dx)}{2(xy)^{1/2}}

What is the first derivative of (xy^{1/2}) ?

The first derivative of (xy^{1/2}) is (y+x\frac{dy)/(dx)}{2(xy)^{1/2}}

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\frac{d}{d x} ((x y)^{\frac{1}{2}})

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

Solution steps

\frac{d}{d x} ((x y)^{\frac{1}{2}})

Apply the chain rule:

\frac{1}{2 ( x y ) ^{\frac{1}{2}}} \frac{d}{d x} (x y)

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

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

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

Simplify

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

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

x \to 0 lim (\frac{tan ( x )}{6 x})

- 5 cos (5 t)

x y \frac{d y}{d x} = x^{2} - y^{2}

f (x) = x + 5, at x = 4

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

What is the derivative of (xy^{1/2}) ?
The derivative of (xy^{1/2}) is (y+x\frac{dy)/(dx)}{2(xy)^{1/2}}
What is the first derivative of (xy^{1/2}) ?
The first derivative of (xy^{1/2}) is (y+x\frac{dy)/(dx)}{2(xy)^{1/2}}