How do you implicitly differentiate #-3=cos(y-x)/x#?

Answer 1

#frac{"d"y}{"d"x} = 1 - cot(y-x)/x#

Differentiate both sides w.r.t. #x#. Use the quotient rule and the chain rule along the way.
#frac{"d"}{"d"x}(3) = frac{"d"}{"d"x}( cos(y-x)/x )#
#0 = frac{ xfrac{"d"}{"d"x}( cos(y-x) ) - cos(y-x)frac{"d"}{"d"x}(x) }{x^2}#
#0 = x(-sin(y-x))frac{"d"}{"d"x}(y-x) - cos(y-x)#
#0 = xsin(y-x)(frac{"d"y}{"d"x}-1) + cos(y-x)#
Now, we just have to make #frac{"d"y}{"d"x}# the subject of formula. Begin by bringing all the terms containing #frac{"d"y}{"d"x}# to one side.
#xsin(y-x) - cos(y-x) = xsin(y-x)frac{"d"y}{"d"x}#
#frac{"d"y}{"d"x} = frac{xsin(y-x) - cos(y-x)}{xsin(y-x)}#
#= 1 - cot(y-x)/x#
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Answer 2

To implicitly differentiate (-3 = \frac{\cos(y-x)}{x}):

Differentiate both sides of the equation with respect to (x), using the quotient rule on the right side.

The derivative of (-3) with respect to (x) is (0).

For the right side, apply the quotient rule:

Let ( u = \cos(y-x) ) and ( v = x ).

Then, ( u' = -\sin(y-x)(y'-x') ) and ( v' = 1 ).

Using the quotient rule, the derivative of (\frac{u}{v}) is:

[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} ]

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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.

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.

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.

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.

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