How do you differentiate #g(x) = sqrt(x^2-3)sec(2-x)# using the product rule?

Question

1321 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-12T22:58:25

#g'(x)=sec(2-x)(x-(x^2-3)tan(2-x))/(sqrt(x^2-3))#

#(df(x)h(x))/(dx)=f'(x)h(x)+f(x)h'(x)#

Here

#f(x)=sqrt(x^2-3)#

#f'(x)=(2x)/(2sqrt(x^2-3))=x/(sqrt(x^2-3))# (by the chain rule), and

#h(x)=sec(2-x)#

#h'(x)=-tan(2-x)sec(2-x)# (by the chain rule).

Therefore,

#g'(x)=(xsec(2-x))/(sqrt(x^2-3))-sqrt(x^2-3)tan(2-x)sec(2-x)#

#=sec(2-x)(x/(sqrt(x^2-3))-sqrt(x^2-3)tan(2-x))#

#=sec(2-x)(x-(x^2-3)tan(2-x))/(sqrt(x^2-3))#