How do you differentiate $y = arccos (x^{7} + \sqrt{5})$ $y= arccos( x^7+sqrt5 )$ ?

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Answer 1 · 2015-05-18T19:58:16

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 $u$ ,

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

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

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