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

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Answer 1 · 2017-04-27T19:49:56

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

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#