# Find the derivative using first principles? : sin sqrt(x)

Feb 9, 2017

Feb 9, 2017

$\frac{d}{\mathrm{dx}} \sin \sqrt{x} = \frac{\cos \sqrt{x}}{2 \sqrt{x}}$

#### Explanation:

By definition of the derivative:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

So with $f \left(x\right) = \sin \sqrt{x}$ we have;

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin \sqrt{x + h} - \sin \left(\sqrt{x}\right)}{h}$

Using $\sin A - \sin B \equiv 2 \sin \left(\frac{1}{2} \left(A - B\right)\right) \cos \left(\frac{1}{2} \left(A + B\right)\right)$ we can reqrite this as;

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{2 \sin \left(\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)\right) \cos \left(\frac{1}{2} \left(\sqrt{x + h} + \sqrt{x}\right)\right)}{h}$

We now use a little trick as ;

$\left(\sqrt{x + h} - \sqrt{x}\right) \left(\sqrt{x + h} + \sqrt{x}\right) = {\left(\sqrt{x + h}\right)}^{2} - {\left(\sqrt{x}\right)}^{2}$
$\text{ } = x + h - x$
$\text{ } = h$

And so we can replace $h$ in the denominator as follows:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{2 \sin \left(\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)\right) \cos \left(\frac{1}{2} \left(\sqrt{x + h} + \sqrt{x}\right)\right)}{\left(\sqrt{x + h} - \sqrt{x}\right) \left(\sqrt{x + h} + \sqrt{x}\right)}$
$\setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin \left(\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)\right) \cos \left(\frac{1}{2} \left(\sqrt{x + h} + \sqrt{x}\right)\right)}{\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right) \left(\sqrt{x + h} + \sqrt{x}\right)}$
$\setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin \left(\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)\right)}{\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)} \cdot \cos \frac{\frac{1}{2} \left(\sqrt{x + h} + \sqrt{x}\right)}{\sqrt{x + h} + \sqrt{x}}$
$\setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin \left(\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)\right)}{\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)} \cdot {\lim}_{h \rightarrow 0} \cos \frac{\frac{1}{2} \left(\sqrt{x + h} + \sqrt{x}\right)}{\sqrt{x + h} + \sqrt{x}}$

Let's look at the first limit;

${\lim}_{h \rightarrow 0} \frac{\sin \left(\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)\right)}{\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)}$

If we put $\theta = \frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)$ then $\theta \rightarrow 0$ as $h \rightarrow 0$, and so;

${\lim}_{h \rightarrow 0} \frac{\sin \left(\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)\right)}{\frac{1}{2} \left(\sqrt{x + h} - \sqrt{x}\right)} = {\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta}$

Which is a standard trig calculus limit, and is equal to unity.

And so now we have:

$f ' \left(x\right) = 1 \cdot {\lim}_{h \rightarrow 0} \cos \frac{\frac{1}{2} \left(\sqrt{x + h} + \sqrt{x}\right)}{\sqrt{x + h} + \sqrt{x}}$
$\setminus \setminus = \cos \frac{\frac{1}{2} \left(\sqrt{x} + \sqrt{x}\right)}{\sqrt{x} + \sqrt{x}}$
$\setminus \setminus = \cos \frac{\sqrt{x}}{2 \sqrt{x}}$