How do you differentiate #f(x)=x^2lnsec(x^2)# using the chain rule?

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Answer 1 · 2015-11-14T12:52:37

#dy/dx =2x(x^2*tan(x^2) + ln(sec(x^2)))#

So we have

#y = x^2ln(sec(x^2))#

Derivating and using the product rule we have

#dy/dx = x^2d/dx(ln(sec(x^2)) + ln(sec(x^2))d/dxx^2#

#dy/dx = x^2d/dx(ln(sec(x^2)) + 2xln(sec(x^2))#

Let's say #sec(x^2) = u# then, by the chain rule we have

#dy/dx = x^2d/dx(ln(u)) + 2xln(sec(x^2))#
#dy/dx = x^2d/(du)(ln(u))(du)/dx + 2xln(sec(x^2))#
#dy/dx = x^2/u*d/dx(sec(x^2)) + 2xln(sec(x^2))#

Let's say #x^2 = v# then we have

#dy/dx = x^2/u*d/(dv)sec(v)*(dv)/dx + 2xln(sec(x^2))#

#dy/dx = x^2/u*tan(v)*sec(v)*2x + 2xln(sec(x^2))#

Putting it all in terms of #x# we have

#dy/dx =2x(x^2/sec(x^2)*tan(x^2)*sec(x^2) + ln(sec(x^2)))#

#dy/dx =2x(x^2*tan(x^2) + ln(sec(x^2)))#