I am having trouble with this questoin... Find the slope (exact value) of the normal to the curve xe^y + ylnx = 2 at the point (1, ln2). Thanks for any help!
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Originally Posted by ty2391 I am having trouble with this questoin... Find the slope (exact value) of the normal to the curve xe^y + ylnx = 2 at the point (1, ln2). Thanks for any help! Differentiate both sides of with respect to x, using the product rule as required: . Note: From the chain rule, the derivative with respect to x of is . Now substitute x = 1 and y = ln 2. Note: . Now solve for .
I ended up with: 2 / ln 2 + 2 as my answer... Can someone confirm this?
Originally Posted by ty2391 I ended up with: 2 / ln 2 + 2 as my answer... Can someone confirm this? . Substitute x = 1 and y = ln 2: ....
It's looking for the normal though, so I guess I am right?
Originally Posted by ty2391 It's looking for the normal though, so I guess I am right? (My boldface). After what I posted, you need to guess!?
Last edited by mr fantastic; Feb 22nd 2008 at 01:24 PM.
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