How do you use the chain rule to differentiate #(sinx)^10#?

Question

2678 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-05T08:23:54

#(dy)/(dx)=10sin^9xcosx#

By treating #(sinx)^10# as a function in terms of #sinx#, and #sinx# as a function in terms of #x#, chain rule can be applied where:

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

Let #u=sinx#

#:.y=u^10#

#(dy)/(du)=10u^9=10(sinx)^9#

#(du)/(dx)=cosx->#the derivative of #sinx# is #cosx#

#:.(dy)/(dx)=10(sinx)^9*cosx=10sin^9xcosx#