# How do you find the derivative of y=tan^2(5x) ?

Jul 24, 2014

The derivative of tan2(5x) is 10tan(5x)sec2(5x).

#### Explanation:

The derivative of ${\tan}^{2} \left(5 x\right)$ is $10 \tan \left(5 x\right) {\sec}^{2} \left(5 x\right)$.

Our function is the composite of three simpler functions:
Start with $x$ and multipy it by $5$.
Then find the tangent of that.
Finally, find the square of that.

So to find the derivative we need the threefold chain rule:

If $k \left(x\right) = f \left(g \left(h \left(x\right)\right)\right)$, then
k'(x) = f'((g(h(x))*g'(h(x))*h'(x).

In our problem, $f \left(x\right) = {x}^{2} , g \left(x\right) = \tan \left(x\right)$, and $h \left(x\right) = 5 x$,
so $f ' \left(x\right) = 2 x , g ' \left(x\right) = {\sec}^{2} \left(x\right)$ and $h ' \left(x\right) = 5$

Thus k'(x) = f'((g(h(x))*g'(h(x))*h'(x)
$= f ' \left(\tan 5 x\right) \cdot g ' \left(5 x\right) \cdot h ' \left(x\right)$
$= 2 \tan \left(5 x\right) \cdot {\sec}^{2} \left(5 x\right) \cdot 5$
$= 10 \tan \left(5 x\right) {\sec}^{2} \left(5 x\right)$

Note: You might prefer doing the problem by letting
$u = 5 x , v = \tan u , y = {v}^{2}$

Then
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{dv}}\right) \left(\frac{\mathrm{dv}}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$
$= 2 v \cdot {\sec}^{2} u \cdot 5$
$= 10 \tan \left(5 x\right) {\sec}^{2} \left(5 x\right)$

Many students prefer the second method, but occasionally you may encounter a problem for which the notation in the second method becomes ambiguous, so it is good to know both methods.