Lucille

Qual è la derivata di # y = sec (x ^ 2) #? #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)#