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

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Answer 1 · 2015-05-18T19:58:16

If #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)#