What is the derivative of 2x^{1/2}y^{1/2} ?

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

What is the first derivative of 2x^{1/2}y^{1/2} ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{\frac{1}{2}}, g = y^{\frac{1}{2}}

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{4}{cos ( x )} + \frac{1}{tan ( x )})

\frac{\partial}{\partial y} (3 x sin (x y))

y^{^{''}} + y = sin (6 t)

y = x^{4} + 2 x^{2} - x, (1, 2)

\int 16 x ln (6 x) d x

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