# How do you differentiate y = tan^3(x^3)?

Jul 27, 2015

Use the chain rule twice.

#### Explanation:

It is important to remember the convention for powers of trigonometric functions:

${\tan}^{3} \left({x}^{3}\right)$ means "raise $x$ to the third power, then find the tangent of that number and finally raise that tangent to the third power". In notation:

${\left(\tan \left({x}^{3}\right)\right)}^{3}$

The outermost function is the (second) cube.
And we know that $\frac{d}{\mathrm{dx}} \left({u}^{3}\right) = 3 {u}^{2} \frac{\mathrm{du}}{\mathrm{dx}}$
In this case $u = \tan \left({x}^{3}\right)$

So we get

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\left(\tan \left({x}^{3}\right)\right)}^{2} \cdot \frac{d}{\mathrm{dx}} \left(\tan \left({x}^{3}\right)\right)$

Now we need the derivative of $\tan \left({x}^{3}\right)$.

We'll use the chain rule again:

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

We get

$\frac{d}{\mathrm{dx}} \left(\tan \left({x}^{3}\right)\right) = {\sec}^{2} \left({x}^{3}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{3}\right)$

$= {\sec}^{2} \left({x}^{3}\right) \cdot 3 {x}^{2}$

Putting it all together, we have:

$y = {\tan}^{3} \left({x}^{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\tan}^{2} \left({x}^{3}\right) \cdot {\sec}^{2} \left({x}^{3}\right) \cdot 3 {x}^{2}$

Which can be written:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 9 {x}^{2} {\tan}^{2} \left({x}^{3}\right) {\sec}^{2} \left({x}^{3}\right)$