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Calculus1 20 Online
OpenStudy (anonymous):

Derive secx-Sqrt(2)tanx I get sinx/(sqrt(2)cos)^2 But it says the answer is sec(x)tan(x)-sqrt(2)sec^2(x) how do I do it?

OpenStudy (zubhanwc3):

alright, lets look at the formula \[\sec x-\sqrt{2}\tan x\]

OpenStudy (zubhanwc3):

moving from the left to the right, the derivative of sec x is Sec x Tan x

OpenStudy (zubhanwc3):

then going to the right, to find the derivative of \[\sqrt{2}\tan x\] we need to use the chain rule

OpenStudy (zubhanwc3):

er my bad, not chain rule, product rule

OpenStudy (zubhanwc3):

say \[\sqrt{2}=u\] and tan x = v

OpenStudy (zubhanwc3):

product rule is uv' + vu'

OpenStudy (zubhanwc3):

now plug it in, derivative of tan x is sec^2 x and the derivative of sqrt(2) is 0 so the derivative of sqrt(2) tan x is \[\sqrt{2}\sec ^{2}x\]

OpenStudy (zubhanwc3):

now to put it together, \[\sec x \tan x-\sqrt{2}\sec ^{2}x\]

OpenStudy (xapproachesinfinity):

differentiate not derive!

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