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Hi, I'm really stuck with a calculus question, I was wondering if you could help me out:
What is dy/dx of e^(xy) = x when x = 2 (Hint: take the natural log of both sides).
I understand what the question is asking me, and I know how to take the ln of both sides but other than that I'm stuck solving the derivative.
Thanks
Charlotte
Taking the ln of both sides gives xy = ln(x). This is where the hint takes you, but there is one more step. The only thing left is to divide both sides by x, and then you have y by itself. Using the quotient rule, the derivative can then be found.
This problem could also be done from the start, giving
e^(xy)d(xy)/dx = 1, so d(xy)/dx = xy' + y, so you'd have
e^(xy)(xy' + y) = 1.
Solving for dy/dx would then involve getting rid of a y and a y', so you'd have to solve for y anyway. This shows that the first paragraph is probably the best approach.
Hope this answer helps you and best of luck. You may write back for further questions on this or any other subject.
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| Comment | Thanks a lot, you really helped out. It was my own stupid mistake and I knew what I was doing wrong in the first line. Intead of having xy when taking the ln, I was putting down ln(xy), defeating the purpose of removing the e. Again, thanks a lot. Charlotte | ||
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