What is the Derivative of #y=sec(x^2)#?
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"How does a vector quantity differ from a scalar quantity?"
1 Answer
Aug 13, 2014
#y'=2x*sec(x^2)tan(x^2)# Solution
let's
#y=f(g(x))# Using Chain Rule, we get
#y'=f'(g(x))*g'(x)# for given problem, which is
#y=sec(x^2)# differentiating with respect to
#x# using Chain Rule,
#y'=sec(x^2)tan(x^2)*(x^2)'#
#y'=sec(x^2)tan(x^2)*2x#
#y'=2x*sec(x^2)tan(x^2)#
