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

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Answer 1 · 2015-12-06T15:21:54

The derivative of secant is #secxtanx# and the derivative of #3-8x " is" -8# ...

#y'=sec(3-8x)xxtan(3-8x)xx-8#

hope that helped

Answer 2 · 2015-12-06T15:24:52

#y'=8sec(8x-3)tan(8x-3)#

According to the chain rule, #d/dx[sec(u)]=u'sec(u)tan(u)#.

Therefore, #d/dx[sec(3-8x)]=d/dx[3-8x]sec(3-8x)tan(3-8x)#.

#d/dx[3-8x]=-8#, so:

#y'=-8sec(3-8x)tan(3-8x)#

Another slightly different approach would be to recognize that #cos(-a)=cos(a)#, and since #sec(a)=1/cos(a),sec(-a)=sec(a)#.

Thus, #sec(3-8x)=sec(8x-3)#, and #y'=8sec(8x-3)tan(8x-3)#.