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

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

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Answer 1 · 2016-07-14T18:18:18

$\frac{d y}{d x} = 3 cos 3 x - 4 sin 4 x$ $(dy)/(dx) = 3cos3x - 4sin4x$

Explanation:

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

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $(dy)/(dx) = (dy)/(du)*(du)/(dx)$

For the first term:

$\frac{d y}{d u} = cos 3 x, \frac{d u}{d x} = 3 \Rightarrow \frac{d y}{d x} = 3 cos 3 x$ $dy/(du) = cos3x, (du)/(dx) = 3 implies dy/(dx) = 3cos3x$

Similarly for the second term:

$\frac{d y}{d u} = - sin 4 x, \frac{d u}{d x} = 4 \Rightarrow \frac{d y}{d x} = - 4 sin 4 x$ $(dy)/(du) = -sin4x, (du)/(dx) = 4 implies (dy)/(dx) = -4sin4x$

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

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