What Did I Do Wrong? Analyzing y = x tan x

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SUMMARY

The discussion centers on the differentiation of the function y = x tan x, specifically the calculation of the second derivative y''. The user initially derived y' as tan x + x sec²(x) and attempted to compute y'' but encountered discrepancies with the expected result. The correct second derivative, as clarified in the discussion, is 2sec²(x)(tan x + 1), highlighting the importance of proper application of the product rule and chain rule in differentiation.

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y = x tan x

y' = tan x + x*sec^2(x)

y'' = sec^2(x) + d/dx (x*sec^2(x))

= ... + (d/dx(sec^2(x)) + sec^2(x))

= ... + (2secx + sec^2(x))

= 2sec^3 x + sec^4(x)

what did I do wrong?? in book it says y'' = something else.
 
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= ... + (2secx + sec^2(x))
First term should have d/dx(secx)=tanx.secx multipled, i.e. first term is 2sec2x.tanx.
 
= ... + (d/dx(sec^2(x)) + sec^2(x))

= ... + (2secx + sec^2(x))

= 2sec^3 x + sec^4(x)

You multiplied the first "sec^2 x" instead of adding!
 
I found the answer this morning,

something like

2sec^2 x(tanx + 1)
 

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