How do you determine #(dy)/(dx)# given #y=sec(sqrt(x^3+x))#?

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Answer 1 · 2018-05-19T08:53:36

#(dy)/(dx)=sec(sqrt(x^3+x))tan(sqrt(x^3+x))(3x^2+1)/(2sqrt(x^3+x))#

We use chain rule here to differentiate. Let #y=sec(g(x))#, where #g(x)=sqrt(h(x))# and #h(x)=x^3+x#

As such #(dy)/(dg(x))=sec(g(x))tan(g(x))#, #(dg(x))/(dh(x))=1/(2sqrt(h(x))# and #(dh(x))/(dx)=3x^2+1#

Hence #(dy)/(dx)=(dy)/(dg(x)) * (dg(x))/(dh(x))*(dh(x))/(dx)#

= #sec(g(x))tan(g(x)) * 1/(2sqrt(h(x))) * (3x^2+1)#

= #sec(sqrt(h(x)))tan(sqrt(h(x))) * 1/(2sqrt(x^3+x)) * (3x^2+1)#

= #sec(sqrt(x^3+x))tan(sqrt(x^3+x)) * 1/(2sqrt(x^3+x)) * (3x^2+1)#

= #sec(sqrt(x^3+x))tan(sqrt(x^3+x))(3x^2+1)/(2sqrt(x^3+x))#