The expression $\frac{\sin \left(\theta \right)}{3}+\frac{c}{\theta }$ contains several variables: the variable θ is chosen for derivation

To ease differentiation, the expression $\frac{c}{\theta }$ is turned into $c{\theta }^{-1}$

Let's perform the differentiation: $\frac{\mathrm{d}}{\mathrm{d}\theta }(\frac{\sin \left(\theta \right)}{3}+c{\theta }^{-1})$

ⓘ The derivative of a sum is the sum of the derivatives: $\frac{\mathrm{d}}{\mathrm{d}\theta }(\frac{\sin \left(\theta \right)}{3}+c{\theta }^{-1})$ = $\frac{\mathrm{d}}{\mathrm{d}\theta }\left(\frac{\sin \left(\theta \right)}{3}\right)+\frac{\mathrm{d}}{\mathrm{d}\theta }\left(c{\theta }^{-1}\right)$

$\frac{\mathrm{d}}{\mathrm{d}\theta }\left(\frac{\sin \left(\theta \right)}{3}\right)+\frac{\mathrm{d}}{\mathrm{d}\theta }\left(c{\theta }^{-1}\right)$

Apply the differentiation rule: $\frac{\mathrm{d}}{\mathrm{d}\theta }\left(f^n\right)=n·\frac{\mathrm{d}}{\mathrm{d}\theta }\left(f\right)·f^{n-1}$ with f = $\theta$

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u'. The derivative of sin(x) is cos(x): $\frac{\mathrm{d}}{\mathrm{d}\theta }\left(\frac{\sin \left(\theta \right)}{3}\right)$ = $\frac{1}{3}·\cos \left(\theta \right)$■For differentiation with respect to $\theta$, $c$ is treated as constant. A constant factor stays in front of the derivative, (k*u)' = k*u': $\frac{\mathrm{d}}{\mathrm{d}\theta }\left(c{\theta }^{-1}\right)$ = $c·\frac{\mathrm{d}}{\mathrm{d}\theta }\left({\theta }^{-1}\right)$■The derivative of x^n is n*x^(n-1): $\frac{\mathrm{d}}{\mathrm{d}\theta }\left({\theta }^{-1}\right)$ = $-1·{\theta }^{-2}$

$\frac{1}{3}·\cos \left(\theta \right)+c·(-1·{\theta }^{-2})$

ⓘ Multiplying a fraction by a term multiplies its numerator: $\frac{1}{3}·\cos \left(\theta \right)$ = $\frac{1·\cos \left(\theta \right)}{3}$ = $\frac{\cos \left(\theta \right)}{3}$■Multiplying by -1 gives the opposite: $c·\left(-1\right)$ = $-c$

$\frac{\cos \left(\theta \right)}{3}-c{\theta }^{-2}$

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base: $c{\theta }^{-2}$ = $\frac{c}{{\theta }^2}$

▶Derivative:   $\frac{\cos \left(\theta \right)}{3}-\frac{c}{{\theta }^2}$