How do you implicitly differentiate #7=1-e^y/(xy)#?

Answer 1

I found: #(dy)/(dx)=y/(x(y-1))#

We need to remember that #y# will be itself a function of #x# so we need to derive it accordingly. For example, if you have #y^2# the derivative will be: #2y*(dy)/(dx)# to consider its dependence from #x#. In our case we have: #0=0-(xye^y*(dy)/(dx)-e^y(y+x(dy)/(dx)))/(xy)^2# rearranging: #(dy)/(dx)[xye^y-xe^y]=ye^y# #(dy)/(dx)=(ye^y)/(xe^y(y-1))=y/(x(y-1))#
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Answer 2

To implicitly differentiate (7 = 1 - \frac{e^y}{xy}) with respect to (x), apply the chain rule and product rule. After differentiating both sides, solve for (\frac{{dy}}{{dx}}).

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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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