How do you use the chain rule to differentiate #y=sin(3x)#?

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Answer 1 · 2016-07-31T17:13:11

#dy/dx=3cos(3x)#.

Let #y=sint, where, t=3x#.

Thus, #y# is a function of #t#, & #t# of #x#.

Then, by the Chain Rule, #dy/dx=dy/(dt)*dt/dx..............(1)#.

#y=sint rArr dy/dt=cost................(2)#.

#t=3x rArr dt/dx=3...........................(3)#.

Then, by #(1),(2), and,(3)#, we have,

#dy/dx=(cost)(3)=3cost=3cos(3x)#.