How do you differentiate # y=sqrt(sec (x^2/pi - xpi))-sec (sqrt(x^2/pi - xpi))# using the chain rule?

1 Answer
Feb 27, 2017

Answer:

#(dy)/(dx)=((2x)/pi-pi)[1/(2sqrt(sec(x^2/pi-xpi)))sec(x^2/pi-xpi)tan(x^2/pi-xpi)-sec(sqrt(x^2/pi-xpi))tan(sqrt(x^2/pi-xpi))xx1/(2sqrt(x^2/pi-xpi))]#

Explanation:

Chain Rule - In order to differentiate a function of a function, say #y, =f(g(x))#, where we have to find #(dy)/(dx)#, we need to do (a) substitute #u=g(x)#, which gives us #y=f(u)#. Then we need to use a formula called Chain Rule, which states that #(dy)/(dx)=(dy)/(du)xx(du)/(dx)#. In fact if we have something like #y=f(g(h(x)))#, we can have #(dy)/(dx)=(dy)/(df)xx(df)/(dg)xx(dg)/(dh)#

Further if #f(x)=u(x)+-v(x)#, #(df)/(dx)=(du)/(dx)+-(dv)/(dx)#

Hence here #(dy)/(dx)=d/(dx)sqrt(sec(x^2/pi-xpi))-d/(dx)sec(sqrt(x^2/pi-xpi))#

Let us workout the two differentials separately and then we can combine them.

#d/(dx)sqrt(g(x))#, where #g(x)=sec(h(x))# and #h(x)=x^2/pi-xpi#

#:.d/(dx)sqrt(sec(x^2/pi-xpi))# using chain rule

= #1/(2sqrt(g(x)))xxsec(h(x))tan(h(x))xx((2x)/pi-pi)#

= #1/(2sqrt(sec(x^2/pi-xpi)))xxsec(x^2/pi-xpi)tan(x^2/pi-xpi)xx((2x)/pi-pi)#

and #d/(dx)sec(sqrt(x^2/pi-xpi))#

= #sec(sqrt(x^2/pi-xpi))tan(sqrt(x^2/pi-xpi))xx1/(2sqrt(x^2/pi-xpi))xx((2x)/pi-pi)#

Hence, #(dy)/(dx)=1/(2sqrt(sec(x^2/pi-xpi)))xxsec(x^2/pi-xpi)tan(x^2/pi-xpi)xx((2x)/pi-pi)-sec(sqrt(x^2/pi-xpi))tan(sqrt(x^2/pi-xpi))xx1/(2sqrt(x^2/pi-xpi))xx((2x)/pi-pi)#

= #((2x)/pi-pi)[1/(2sqrt(sec(x^2/pi-xpi)))sec(x^2/pi-xpi)tan(x^2/pi-xpi)-sec(sqrt(x^2/pi-xpi))tan(sqrt(x^2/pi-xpi))xx1/(2sqrt(x^2/pi-xpi))]#