Question

What is the derivative of `tan^10( 5x)`?

Answer 1

`(dy)/(dx)=50tan^9 5xsec^2 5x`

Explanation:

This needs the use of the chain rule twice

`y=tan^10(5x)`

`"let " u=5x----(1)`

then

`y=tan^10u`

`"let " v=tanu=>y=v^10`

`(dy)/(du)=(dy)/(dv)xx(dv)/(du)`

`(dy)/(dv)=10v^9`

`(dv)/(du)=sec^2u`

`:.(dy)/(du)=10v^9xxsec^2u`

`:.(dy)/(du)=10tan^9usec^2u`

we want`(dy)/(dx)`

`(dy)/(dx)=(dy)/(du)xx(du)/(dx)`

from `(1)`

`(du)/(dx)=5`

substituting back in and substituting the `u" function back"`

`(dy)/(dx)=5xx10tan^9 5xsec^2 5x`

`(dy)/(dx)=50tan^9 5xsec^2 5x`