Solve beta=2pi*f/u

Evaluate

β = 2 π \times \frac{f}{u}

Simplify

β = \frac{2 π f}{u}

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

\frac{\partial β}{\partial f} = \frac{\partial}{\partial f} (\frac{2 π f}{u})

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 β}{\partial f} = \frac{\frac{\partial}{\partial f} ( 2 π f ) u - 2 π f \times \frac{\partial}{\partial f} ( u )}{u ^{2}}

Evaluate

More Steps

\frac{\partial β}{\partial f} = \frac{2 π u - 2 π f \times \frac{\partial}{\partial f} ( u )}{u ^{2}}

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

\frac{\partial β}{\partial f} = \frac{2 π u - 2 π f \times 0}{u ^{2}}

Any expression multiplied by 0 equals 0

\frac{\partial β}{\partial f} = \frac{2 π u - 0}{u ^{2}}

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

\frac{\partial β}{\partial f} = \frac{2 π u}{u ^{2}}

Solution

More Steps

\frac{\partial β}{\partial f} = \frac{2 π}{u}

Function