How do you take the derivative of # y=tan^2(x^3)#?

Question

Ezra Adams · Accepted Answer

#y^' = 6x^2 * tan(x^3) * sec^2(x^3)#

You can differentiate this function by using the chain rule twice, once for #u^2#, with #u = tan(x^3)#, and once more for #tan(t)#, with #t = x^3#.

You'll get this

#d/dx(y) = d/(du)u^2 * d/dx(u)#

#y^' = 2u * d/dxtan(x^3)#

The derivative of #tanx^3# will be equal to

#d/dx(tant) = d/(dt)tan(t) * d/(dx)(t)#

#d/dx(tant) = sec^2t * d/dx(x^3)#

#d/dx(tan(x^3)) = sec^2(x^3) * 3x^2#

As a result, your target derivative will equal

#y^' = 2 * tan(x^3) * 3x^2 * sec^2(x^3)#

#y^' = color(green)(6x^2 * tan(x^3) * sec^2(x^3))#

Ezra Adams · Answer

undefined To take the derivative of $y = \tan^2(x^3)$, you can use the chain rule and the derivative of the tangent function:

1. Apply the chain rule: 
$$\frac{d}{dx} \tan^2(u) = 2\tan(u) \cdot \frac{d}{dx} \tan(u)$$

2. Let $u = x^3$, so $y = \tan^2(u)$.

3. Find $\frac{d}{du} \tan(u)$ and $\frac{d}{dx} x^3$.
   - $\frac{d}{du} \tan(u) = \sec^2(u) \cdot \frac{d}{du} u$
   - $\frac{d}{dx} x^3 = 3x^2$

4. Substitute back:
   $$\frac{d}{dx} \tan^2(x^3) = 2\tan(x^3) \cdot \sec^2(x^3) \cdot 3x^2$$

So, the derivative of $y = \tan^2(x^3)$ is:
$$y' = 6x^2 \tan(x^3) \sec^2(x^3)$$