# How do you differentiate  y=sec^2 (3 - 8x) using the chain rule?

Dec 28, 2015

Remember that ${\sec}^{2} \left(u\right) = {\left(\sec \left(u\right)\right)}^{2}$

#### Explanation:

So, for y=sec^2(3-8x)) we 'really' have

$y = {\left(\sec \left(3 - 8 x\right)\right)}^{2}$

The outermost function is the square function. Applying the chain rule, we get

$y ' = \left[2 {\left(\sec \left(3 - 8 x\right)\right)}^{1}\right] \frac{d}{\mathrm{dx}} \left(\sec \left(3 - 8 x\right)\right)$

Now, we use the chain rule the second time, to see:

$\frac{d}{\mathrm{dx}} \left(\sec \left(3 - 8 x\right)\right) = \left[\sec \left(3 - 8 x\right) \tan \left(3 - 8 x\right)\right] \frac{d}{\mathrm{dx}} \left(3 - 8 x\right)$

And, of course $\frac{d}{\mathrm{dx}} \left(3 - 8 x\right) = - 8$

For y=sec^2(3-8x)), we get

$y ' = 2 \sec \left(3 - 8 x\right) \sec \left(3 - 8 x\right) \tan \left(3 - 8 x\right) \left(- 8\right)$.

Finally, simplify to get

$y ' = - 16 {\sec}^{2} \left(3 - 8 x\right) \tan \left(3 - 8 x\right)$