# Question #8b5f0

Oct 16, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(x\right)$

#### Explanation:

What we will need:

• ${\lim}_{x \to 0} \sin \frac{x}{x} = 1$

This can be derived using the squeeze theorem together with some geometric arguments.

• $\left\{\begin{matrix}{\lim}_{x \to a} f \left(x\right) = {L}_{f} \\ {\lim}_{x \to a} g \left(x\right) = {L}_{g}\end{matrix}\right. \implies {\lim}_{x \to a} f \left(x\right) g \left(x\right) = {L}_{f} {L}_{g}$

It can be shown using the definition of a limit that if two functions have finite limits at a point, then the limit of the sum of those functions is the sum of their limits at that point.

• $\left\{\begin{matrix}{\lim}_{x \to a} f \left(x\right) = {L}_{f} \\ {\lim}_{x \to a} g \left(x\right) = {L}_{g}\end{matrix}\right. \implies {\lim}_{x \to a} \left(f \left(x\right) + g \left(x\right)\right) = {L}_{f} + {L}_{g}$

A similar property applies to sums.

• ${\lim}_{x \to 0} \frac{\cos \left(x\right) - 1}{x} = 0$

This can be derived from the first limit shown above by using some algebraic manipulation to show that $\frac{\cos \left(x\right) - 1}{x} = - \left(\sin \frac{x}{x}\right) \left(\sin \frac{x}{1 + \cos \left(x\right)}\right)$ and using the multiplication property shown above.

• $\frac{d}{\mathrm{dx}} f \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

The definition of a derivative.

With the above, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \sin \left(x\right)$

$= {\lim}_{h \to 0} \frac{\sin \left(x + h\right) - \sin \left(x\right)}{h}$

$= {\lim}_{h \to 0} \frac{\sin \left(x\right) \cos \left(h\right) + \cos \left(x\right) \sin \left(h\right) - \sin \left(x\right)}{h}$

$= {\lim}_{h \to 0} \frac{\sin \left(x\right) \left(\cos \left(h\right) - 1\right) + \cos \left(x\right) \sin \left(h\right)}{h}$

$= {\lim}_{h \to 0} \left(\sin \left(x\right) \frac{\cos \left(h\right) - 1}{h} + \cos \left(x\right) \sin \frac{h}{h}\right)$

$= {\lim}_{h \to 0} \sin \left(x\right) \frac{\cos \left(h\right) - 1}{h} + {\lim}_{h \to 0} \cos \left(x\right) \sin \frac{h}{h}$

$= \sin \left(x\right) \cdot 0 + \cos \left(x\right) \cdot 1$

$= \cos \left(x\right)$