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

Question

2099 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-28T18:47:59

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

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

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

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

$y' = [2(sec(3-8x))^1] d/dx(sec(3-8x))$

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

$d/dx(sec(3-8x)) = [sec(3-8x)tan(3-8x)] d/dx(3-8x)$

And, of course $d/dx(3-8x) = -8$

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

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

Finally, simplify to get

$y' = -16sec^2(3-8x)tan(3-8x)$

How do you differentiate y=sec^2 (3 - 8x)y=sec2(3−8x) y=sec^2 (3 - 8x) using the chain rule?