Solve \omega = (g/l)½

Evaluate

ω = \frac{g}{l} \times \frac{1}{2}

Simplify

More Steps

ω = \frac{g}{2 l}

Find the first partial derivative by treating the variable l as a constant and differentiating with respect to g

\frac{\partial ω}{\partial g} = \frac{\partial}{\partial g} (\frac{g}{2 l})

Use differentiation rule \frac{\partial}{\partial x} (\frac{f ( x )}{g ( x )}) = \frac{\frac{\partial}{\partial x} ( f ( x )) \times g ( x ) - f ( x ) \times \frac{\partial}{\partial x} ( g ( x ) )}{( g ( x ) ) ^{2}}

\frac{\partial ω}{\partial g} = \frac{\frac{\partial}{\partial g} ( g ) \times 2 l - g \times \frac{\partial}{\partial g} ( 2 l )}{( 2 l ) ^{2}}

Use \frac{\partial}{\partial x} x^{n} = n x^{n - 1} to find derivative

\frac{\partial ω}{\partial g} = \frac{1 \times 2 l - g \times \frac{\partial}{\partial g} ( 2 l )}{( 2 l ) ^{2}}

Use \frac{\partial}{\partial x} (c) = 0 to find derivative

\frac{\partial ω}{\partial g} = \frac{1 \times 2 l - g \times 0}{( 2 l ) ^{2}}

Any expression multiplied by 1 remains the same

\frac{\partial ω}{\partial g} = \frac{2 l - g \times 0}{( 2 l ) ^{2}}

Any expression multiplied by 0 equals 0

\frac{\partial ω}{\partial g} = \frac{2 l - 0}{( 2 l ) ^{2}}

Evaluate

More Steps

\frac{\partial ω}{\partial g} = \frac{2 l - 0}{4 l ^{2}}

Removing 0 doesn't change the value,so remove it from the expression

\frac{\partial ω}{\partial g} = \frac{2 l}{4 l ^{2}}

Solution

More Steps

\frac{\partial ω}{\partial g} = \frac{1}{2 l}

Function