How do you use the chain rule to differentiate #y=sin^3x+cos^3x#?

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Answer 1 · 2016-10-07T18:11:43

Please follow the instructions below...

#y=sin^3x+cos^3x#

#(dy)/(dx)=d/(dx)(sin^3x)+d/(dx)(cos^3x)#

Now we'll be using the chain rule to get #(dy)/(dx)#...

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

So, here it goes...

#d/(du)(u^3)=3u^2=3sin^2x#

#d/(dx)(sinx)=cosx#

#d/(du)(u^3)*d/(dx)(sinx)=3sin^2xcosx#

Which means that:

#d/(dx)(sin^3x)=3sin^2xcosx#

Now...

#d/(dp)(p^3)=3p^2=3cos^2x#

#d/(dx)(cosx)=-sinx#

#d/(dp)(p^3)*d/(dx)(cosx)=-3cos^2xsinx#

Which means that...

#d/(dx)(cos^3x)=-3cos^2xsinx#

Since...

#(dy)/(dx)=d/(dx)(sin^3x)+d/(dx)(cos^3x)#

#(dy)/(dx)=3sin^2xcosx-3cos^2xsinx#

#=3sinxcosx(sinx-cosx)#

You can then again transform this result if you wish to.