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## Homework Statement

*Given y=xtanx, find y'' (second derivative)*

## Homework Equations

Uh... I'm not even sure if I'm using the right one....

d/dx(tanx) = sec^2x

## The Attempt at a Solution

y=xtanx

y'= (x)(sec^2(x)) + (tanx)(1)

y'= xsec^2(x) + tanx

y'' = [(x)(2sec^3(x)) + sec^2(x)(1)] + sec^2x

y'' = 2xsec^3(x) + sec^2(x) + sec^2(x)

....

The key answer says it is supposed to be (2cosx + 2xsinx)/(cos^3(x))