Differentiation of $\sin v$

Let $y = \sin v$. By General Rule, §4.7, considering $v$ as the independent variable, we have

• FIRST STEP. $y + \Delta y = \sin(v + \Delta v)$.

• SECOND STEP^5.85.9
  $$\Delta y = \sin(v + \Delta v) - \sin v = 2 \cos \left ( v + \frac{\Delta v}{2} \right ) \cdot \sin \frac{\Delta v}{2}.$$

• THIRD STEP.
  $$\frac{\Delta y}{\Delta v} = \cos \left ( v + \frac{\Delta v}{2} \right ) \left ( \frac{\sin \frac{\Delta v}{2}}{\frac{\Delta v}{2}} \right ).$$

• FOURTH STEP. $\frac{dy}{dx} = \cos v$.

  (Since $\lim_{\Delta v \to 0} \left ( \frac{\sin \frac{\Delta v}{2}}{\frac{\Delta v}{2}} \right ) = 1$, by §3.10, and $\lim_{\Delta v \to 0} \cos \left ( v + \frac{\Delta v}{2} \right ) = \cos v$.)

Since $v$ is a function of $x$ and it is required to differentiate $\sin v$ with respect to $x$, we must use formula (A), §5.11, for differentiating a function of a function, namely,

$$\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}.$$

Substituting value $\frac{dy}{dx}$ from Fourth Step, we get $\frac{dy}{dx} = \cos v \frac{dv}{dx}$. Therefore,

$$\frac{d}{dx} (\sin v) = \cos v \frac{dv}{dx}$$

(equation (XI) in §5.1 above).

The statement of the corresponding rules will now be left to the student.