×

# How do you find the derivative of y=tan(2x)?

Aug 5, 2014

Essentially we use the rule:

$\frac{d}{\mathrm{dx}} \left(a \cdot \tan \left(k \cdot x + h\right)\right) = \frac{a \cdot k}{{\cos}^{2} \left(k \cdot x + h\right)}$

or

$a \cdot k \cdot {\sec}^{2} \left(k \cdot x + h\right)$

I'll go through the steps behind it as well:

Derivitave of $y = \tan \left(2 x\right)$

The first step is to separate it into $\sin \left(x\right)$ and $\cos \left(x\right)$:

$y = \sin \frac{2 x}{\cos} \left(2 x\right)$

Then we integrate using the quotient rule:

$f \left(x\right) = g \frac{x}{h \left(x\right)} ,$ $f ' \left(x\right) = \frac{g ' \left(x\right) \cdot h \left(x\right) - h ' \left(x\right) \cdot g \left(x\right)}{h \left(x\right)} ^ 2$

$y ' = \frac{\cos \left(2 x\right) \times 2 \cos \left(2 x\right) - \sin \left(2 x\right) \times - 2 \sin \left(2 x\right)}{{\cos}^{2} \left(2 x\right)}$

$y ' = \frac{2 {\cos}^{2} \left(2 x\right) + 2 {\sin}^{2} \left(2 x\right)}{{\cos}^{2} \left(2 x\right)}$

$y ' = \frac{2 \left({\cos}^{2} \left(2 x\right) + {\sin}^{2} \left(2 x\right)\right)}{{\cos}^{2} \left(2 x\right)}$

To simplify this further we use the Pythagorean Identity:

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

So $2 \left({\cos}^{2} \left(2 x\right) + {\sin}^{2} \left(2 x\right)\right)$ becomes $2 \times 1 = 2$

Thus:

$y ' = \frac{2 \left({\cos}^{2} \left(2 x\right) + {\sin}^{2} \left(2 x\right)\right)}{{\cos}^{2} \left(2 x\right)}$

becomes

$y ' = \frac{2}{{\cos}^{2} \left(2 x\right)}$

$= 2 {\sec}^{2} \left(2 x\right)$

Jun 20, 2015

There is a simple answer to this.

$\frac{d}{\mathrm{dx}} \left[\tan u\right] = {\sec}^{2} u \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

Thus, with $u = 2 x$ (thus, $\frac{u}{x} = 2$), $\frac{\mathrm{du}}{\mathrm{dx}} = 2$, and

$\frac{d}{\mathrm{dx}} \left[\tan \left(2 x\right)\right] = {\sec}^{2} \left(2 x\right) \cdot 2 = \textcolor{b l u e}{2 {\sec}^{2} \left(2 x\right)}$

We can Use Chain rule for this.

#### Explanation:

We have to find derivative of $\textcolor{b l u e}{y = \tan \left(2 x\right)}$
Let us assume $u = 2 x$
Then
color(blue)(y=tanu & color(blue)(u=2x
By Chain Rule of Differenciation
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$\frac{d}{\mathrm{dx}} \left(\tan 2 x\right) = \left(\frac{d}{\mathrm{du}} \left(\tan u\right)\right) \left(\frac{d}{\mathrm{dx}} \left(2 x\right)\right)$
$\frac{d}{\mathrm{dx}} \left(\tan 2 x\right) =$${\sec}^{2} u \cdot 2 \to 1$
Substitute u=2x in Eqn 1
Therefore
$\frac{d}{\mathrm{dx}} \left(\tan 2 x\right) = 2 {\sec}^{2} 2 x$