Calculus Examples

$\frac{f M}{q} + g M + \frac{k q}{2}$

By the Sum Rule, the derivative of $\frac{f M}{q} + g M + \frac{k q}{2}$ with respect to $f$ is $\frac{d}{d f} [\frac{f M}{q}] + \frac{d}{d f} [g M] + \frac{d}{d f} [\frac{k q}{2}]$ .

$\frac{d}{d f} [\frac{f M}{q}] + \frac{d}{d f} [g M] + \frac{d}{d f} [\frac{k q}{2}]$

Evaluate $\frac{d}{d f} [\frac{f M}{q}]$ .

Tap for more steps...

Since $\frac{M}{q}$ is constant with respect to $f$ , the derivative of $\frac{f M}{q}$ with respect to $f$ is $\frac{M}{q} \frac{d}{d f} [f]$ .

$\frac{M}{q} \frac{d}{d f} [f] + \frac{d}{d f} [g M] + \frac{d}{d f} [\frac{k q}{2}]$

Differentiate using the Power Rule which states that $\frac{d}{d f} [f^{n}]$ is $n f^{n - 1}$ where $n = 1$ .

$\frac{M}{q} \cdot 1 + \frac{d}{d f} [g M] + \frac{d}{d f} [\frac{k q}{2}]$

Multiply $\frac{M}{q}$ by $1$ .

$\frac{M}{q} + \frac{d}{d f} [g M] + \frac{d}{d f} [\frac{k q}{2}]$

Differentiate using the Constant Rule.

Tap for more steps...

Since $g M$ is constant with respect to $f$ , the derivative of $g M$ with respect to $f$ is $0$ .

$\frac{M}{q} + 0 + \frac{d}{d f} [\frac{k q}{2}]$

Since $\frac{k q}{2}$ is constant with respect to $f$ , the derivative of $\frac{k q}{2}$ with respect to $f$ is $0$ .

$\frac{M}{q} + 0 + 0$

Combine terms.

Tap for more steps...

Add $\frac{M}{q}$ and $0$ .

$\frac{M}{q} + 0$

Add $\frac{M}{q}$ and $0$ .

$\frac{M}{q}$