What is the differential dy of y=sec(x^2-1) ?

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Answer 1 · 2018-03-21T00:13:02

#y'=2xsec(x^2-1)tan(x^2-1)#

We'll need the chain rule for this one:

We have...

#y=sec(x^2-1)#

By the chain rule...

#color(blue)(d/dx[sec(x^2-1)]=y'=f'(g(x))*g'(x)#

...and letting

#f(x)=sec(x)# and #g(x)=x^2-1#

So...

#f'(x)=sec(x)tan(x)#

#g'(x)=2x#

Thus...

#y'=sec(x^2-1)tan(x^2-1)*2x#