What is (\partial)/(\partial c)(c^{1/2}+l^{1/2}) ?

The answer to (\partial)/(\partial c)(c^{1/2}+l^{1/2}) is 1/(2c^{1/2)}

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(\partial)/(\partial c)(c^{1/2}+l^{1/2})

\frac{\partial}{\partial c} (c^{\frac{1}{2}} + l^{\frac{1}{2}})

\frac{1}{2 c ^{\frac{1}{2}}}

Solution steps

\frac{\partial}{\partial c} (c^{\frac{1}{2}} + l^{\frac{1}{2}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial c} (c^{\frac{1}{2}}) + \frac{\partial}{\partial c} (l^{\frac{1}{2}})

\frac{\partial}{\partial c} (c^{\frac{1}{2}}) = \frac{1}{2 c ^{\frac{1}{2}}}

\frac{\partial}{\partial c} (l^{\frac{1}{2}}) = 0

= \frac{1}{2 c ^{\frac{1}{2}}} + 0

Simplify

= \frac{1}{2 c ^{\frac{1}{2}}}

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

x \to \infty lim (x + 1)

\frac{\partial}{\partial x} (- k p (x) gx)

y = arcsin (x) - 1 - x^{2}

\frac{d ^{2}}{d x ^{2}} (\frac{x}{x + 7})

What is (\partial)/(\partial c)(c^{1/2}+l^{1/2}) ?
The answer to (\partial)/(\partial c)(c^{1/2}+l^{1/2}) is 1/(2c^{1/2)}