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:

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

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

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