How do you differentiate #y= sin 3x + cos 4x#?

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Answer 1 · 2016-07-14T18:18:18

#(dy)/(dx) = 3cos3x - 4sin4x#

We need to use the chain rule here. It states that for a function #y(u(x))#

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

For the first term:

#dy/(du) = cos3x, (du)/(dx) = 3 implies dy/(dx) = 3cos3x#

Similarly for the second term:

#(dy)/(du) = -sin4x, (du)/(dx) = 4 implies (dy)/(dx) = -4sin4x#

#therefore (dy)/(dx) = 3cos3x - 4sin4x#