Error in "Derivatives of Inverse Functions"?
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Yangi maqolaSharhlar: 3
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Hi James,
x = e^y is indeed the inverse function of y = e^x. To get an inverse function, all you have to do is swap the two variables. You generally solve for the resulting "y", but you don't necessarily have to do that. If you solve x = e^y for y, you get y = lnx, which is the inverse of y = e^x.
If you could link me to the video that you were watching, I'd be happy to verify whether or not there is an error in the video.
Best,
Evan
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Hi Evan,
Here’s the video link.4:45, middle of page. There's a snapshot below.The video is, on its face, appears wrong. It says:y=e^x and x=e^y, right below the two domain range circles.When I put e^8 into my calc I get 2980. x is 8 and y is computed to be 2980. Okay. Now I put e^2980 (or y equals 8) into my calc (that’s y) and I get x equals infinity. Infinity does not equal 8.Hope this helps.Jim
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I can definitely understand why you're confused. What you stated is not correct, because you haven't switched the x and y values. Remember, inverse functions have opposite domains and ranges.
When you plug "a" into a function and you get "b", you should get "a" from plugging "b" into the inverse of that function.
So, if y = e^x is the given function, then e^8 = 2980. Here, x = 8 and y = 2980.
The inverse function, without solving for y, is x = e^y. However, since this is the inverse function, the domain and range has been switched. Now, y = 8 when x = 2980. And, 2980 = e^8 is a true statement.
Does that help?
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