Solve \sigma =n⋅p⋅(1-p)

Simplify

σ = n p (1 - p)

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

\frac{\partial σ}{\partial n} = \frac{\partial}{\partial n} (n p (1 - p))

Use differentiation rule \frac{\partial}{\partial x} (f (x) \times g (x)) = \frac{\partial}{\partial x} (f (x)) \times g (x) + f (x) \times \frac{\partial}{\partial x} (g (x))

\frac{\partial σ}{\partial n} = \frac{\partial}{\partial n} (n p) (1 - p) + n p \times \frac{\partial}{\partial n} (1 - p)

Evaluate

More Steps

\frac{\partial σ}{\partial n} = p (1 - p) + n p \times \frac{\partial}{\partial n} (1 - p)

Evaluate

\frac{\partial σ}{\partial n} = p - p^{2} + n p \times \frac{\partial}{\partial n} (1 - p)

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

\frac{\partial σ}{\partial n} = p - p^{2} + n p \times 0

Evaluate

\frac{\partial σ}{\partial n} = p - p^{2} + 0

Solution

\frac{\partial σ}{\partial n} = p - p^{2}

Function