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

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Answer 1 · 2015-12-28T18:47:59

Remember that #sec^2(u) = (sec(u))^2#

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

# 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)#