How do you differentiate y= arccos( x^7+sqrt5 )y=arccos(x7+5)?

1 Answer
May 18, 2015

If y=arccos(x)y=arccos(x), then (dy)/(dx)=(-x')/(sqrt(1-x^2))

Using the chain rule, we can rename u=x^7+sqrt(5), and, by chain rule definition, find dy/(dx)

dy/(dx)=(dy)/(du)*(du)/(dx)

(dy)/(du)=(-u')/(sqrt(1-u^2))

(du)/(dx)=7x^6

Now,

(dy)/(dx)=(-u')/(sqrt(1-u^2))*7x^6

Substituting u,

(dy)/(dx)=(-7x^6)/(sqrt(1-(x^7-sqrt(5))^2))*7x^6

(dy)/(dx)=(-49x^12)/(sqrt(1-(x^14-2x^7sqrt(5)+5)=(-49x^12)/sqrt(-x^14+2x^7sqrt(5)-4)