# Find the derivative of tan2x from first principles?

Mar 23, 2017

$\frac{d}{\mathrm{dx}} \tan 2 x = 2 {\sec}^{2} 2 x$

#### 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 $f \left(x\right) = \tan 2 x$ we have;

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

Using $\tan A = \sin \frac{A}{\cos} A$ we get

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

Using the identity $\sin \left(A - B\right) \equiv \sin A \cos B - \cos B \sin A$ we get;

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

We now have to rely on some standard limits:

${\lim}_{h \rightarrow 0} \sin \frac{h}{h} = 1 \implies {\lim}_{h \rightarrow 0} \frac{\sin 2 h}{2 h} = 1$

And so we get:

$f ' \left(x\right) = \frac{2}{\cos 2 x \cos 2 x} \cdot 1$
$\text{ } = \frac{2}{{\cos}^{2} 2 x}$
$\text{ } = 2 {\sec}^{2} 2 x$