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How do you differentiate #sin(xy)=1/2#?

Answer 1

Ezra Adams

Here are two ways to do it.

The easy (trick) way

From #sin(xy)=1/2# we conclude that #xy = pi/6 + 2pik# for integer #k#.

So #d/dx(xy) = d/dx(pi/6 + 2pik)#. That is,

#y+x dy/dx = 0#, and

#dy/dx = -y/x#

The non-easy way

#d/dx(sin(xy)) = d/dx(1/2)#

#cos(xy)(d/dx(xy)) = 0#

#cos(xy)(y + x dy/dx) = 0#

Divide by #cos(xy)# (OK, since we know it is not #0#.)

to get

#y+x dy/dx = 0#, and

#dy/dx = -y/x#

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Answer 2

Ezekiel Alexander

To differentiate sin(xy) = 1/2 with respect to x, you would use the chain rule. The derivative would be:

∂/∂x(sin(xy)) = cos(xy) * y.

Similarly, if you want to differentiate with respect to y, it would be:

∂/∂y(sin(xy)) = cos(xy) * x.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.