How do i differentiate $x e^{x y} cos (2 x)$ $xe^(xy)cos(2x)$ with respect to x??

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Answer 1 · 2015-02-05T20:10:12

This is actually a multiplication rule within a multiplication rule.

I'm going to evaluate it as:

$(x e^{x} y) (cos (2 x))$ $(xe^xy)(cos(2x))$

Remember the rule for multiplication:

first(derivative of the 2nd) + second(derivative of the first)

You also need to remember that you must take the derivative of the "inside" of the cos.

$(x e^{x} y) (- 2 sin (2 x)) + (cos (2 x)) (x (e^{x y} y) + e^{x y})$ $(xe^xy)(-2sin(2x))+(cos(2x))(x(e^(xy)y)+e^(xy))$

$- 2 x e^{x y} sin (2 x) + x y e^{x y} cos (2 x) + e^{x y} cos (2 x)$ $-2xe^(xy)sin(2x)+xye^(xy)cos(2x)+e^(xy)cos(2x)$

We can factor out $e^{x y}$ $e^(xy)$

$e^{x y} (- 2 x sin (2 x) + x y cos (2 x) + cos (2 x))$ $e^(xy)(-2xsin(2x)+xycos(2x)+cos(2x))$

How do i differentiate xe^(xy)cos(2x)xexycos(2x)xe^(xy)cos(2x) with respect to x??