Another Implicit Differentiation Question

This forum is such a good resource - I have another question regarding implicit differentiation that I can't find any information in my notes for.

I have the following question...

*Assume x and y are functions of t. Evaluate $\displaystyle \frac {dy}{dt}$ for the following.*

$\displaystyle xe^{y} = 2 - ln(2) + ln(x) ....... [ \frac {dx}{dt}= 6 , x = 2 , y = 0]$

What I'm a little unsure about, is that when working through this, mainly when using the chain rule for exponential functions when trying to find $\displaystyle \frac {dy}{dt}$

I got up to about the second step shown below, but then realized that something isn't right at all. I completed the equation to it's (incorrect) conclusion, but I don't really no how to handle the exponents and Euler's number when working with it.

$\displaystyle \frac {d}{dt} (xe^y) = \frac {d}{dt}(2) - \frac {d}{dt} ln(2) + \frac {d}{dt} ln(x)$

$\displaystyle = [x (ye^y \frac{dy}{dt}) + e^y( \frac{dx}{dt}] = 0 - \frac {0}{2} + \frac {1 \frac {dx}{dt}}{x} $

$\displaystyle = xye^y \frac {dy}{dt} + e^y \frac{dx}{dt} = \frac {1 \frac {dx}{dt}}{x}$

$\displaystyle = xye^y \frac {dy}{dt} = \frac {1 \frac{dx}{dt}} {x} - e^y \frac{dx}{dt}$

=$\displaystyle \frac {dy}{dt} = \frac {1 \frac {dx}{dt}}{x^2ye^y} - \frac {ey\frac {dy}{dx}}{xye^y}$

I'm pretty sure that I've already made a mistake somewhere along the line here, but just to make sure...

$\displaystyle \frac {dy}{dt} = \frac {1(6)}{4)(0)(1)} - \frac {e(0)(6)}{(2)(0)(1)}$

$\displaystyle = \frac {6}{0} - 0$

Which is just ridiculous...

Re: Another Implicit Differentiation Question

Quote:

Originally Posted by

**astuart**

$\displaystyle \frac {d}{dt} (xe^y) = \frac {d}{dt}(2) - \frac {d}{dt} ln(2) + \frac {d}{dt} ln(x)$

$\displaystyle = [x (ye^y \frac{dy}{dt}) + e^y( \frac{dx}{dt}] = 0 - \frac {0}{2} + \frac {1 \frac {dx}{dt}}{x} $

Inside your use of the product rule, it looks like you did this:

$\displaystyle x \frac{d}{dt} (e^y) = x ( y e^y \frac{dy}{dt})$

That's a mistake. That y-factor shouldn't be there (it ends up being the cause of your division by 0). It should be:

$\displaystyle x \frac{d}{dt} (e^y) = x ( e^y \frac{dy}{dt})$

Re: Another Implicit Differentiation Question

Quote:

Originally Posted by

**johnsomeone** Inside your use of the product rule, it looks like you did this:

$\displaystyle x \frac{d}{dt} (e^y) = x ( y e^y \frac{dy}{dt})$

That's a mistake. That y-factor shouldn't be there (it ends up being the cause of your division by 0). It should be:

$\displaystyle x \frac{d}{dt} (e^y) = x ( e^y \frac{dy}{dt})$

Isn't the derivative of $\displaystyle e^y = y(e^y)?$??

Re: Another Implicit Differentiation Question

No. What makes $\displaystyle e^x$ so special is that it's the function that remains unchanged when you take its derivative with respect to x.

$\displaystyle \frac{d}{dx}e^x = e^x$.

What's happening in this problem is the chain rule (which is all that implicit differentiation really is). You're taking a derivative of $\displaystyle e^y$ WITH RESPECT TO TIME t, and so you think of y secretly, *implicitly*, being a function of t, so that y = y(t) (even though you usually don't know exactly what that function is). Then you use the chain rule and this unknown function y(t). It works out like this:

$\displaystyle \frac{d}{dt}e^y = \left( \frac{d}{dy}e^y \right) \left( \frac{d}{dt}y(t) \right)$, so

$\displaystyle \frac{d}{dt}e^y = ( e^y ) \left( \frac{dy}{dt} \right)= e^y \frac{dy}{dt}$.

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A common mental shortcut for remembering the chain rule is to think of dx, du, dt, dy, dw, etc as if they were variables, and then look at this:

$\displaystyle \frac{d}{dt} = \frac{d}{dy} \frac{dy}{dt}$, and $\displaystyle \frac{d}{dv} = \frac{d}{dw} \frac{dw}{dv}$, etc.

Sometimes you have a lot of them:

$\displaystyle \frac{d}{dt} = \frac{d}{dy} \frac{dy}{dw} \frac{dw}{du} \frac{du}{dt}$.

This is just a mnemonic device for remembering how to apply the chain rule. It doesn't technically mean anything (well, it could, but let's avoid that.)

In multivariable calculus, the same thing will apply, with a vengence, and with a lot of summations. But that's later...

Re: Another Implicit Differentiation Question

Thanks for the post, cleared up a couple of things, so it's much appreciated!!