Question

Let y = f(x) be defined by the equation e^xy+ln(y^(2)) = x. Find dy/dx

Answer 1

this requires implicit differentiation. is there a specific problem?

[if it were my problem, i would simplify by writing \(e^{xy}+\ln(y^{2}) = x\) as \(f(x,y) = e^{xy}+2\ln y - x =0 \) and noting that \(y' = -\dfrac{f_x}{f_y}\)

administratively simpler. usually.

Answer 2

@IrishBoy123 I have an answer I'm just not sure how to go about solving this is there any chance you can show me the steps?

Answer 3

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Answer 4

for \(f(x,y) = e^{xy}+2\ln y - x =0\) you have

\[f_x = \dfrac{\partial f}{\partial x} = y e^{xy} -1\]
\[f_y = \dfrac{\partial f}{\partial y} = xe^{xy} + 2 .\frac{1}{y}\]
\[\therefore y' = - \dfrac{y e^{xy} -1}{xe^{xy} + 2 \frac{1}{y}} \\ = - \dfrac{y (ye^{xy} -1)}{xye^{xy} + 2 }\]

are you in that kinda territory?!?!? i just latexed this as i did it.....

Answer 5

@IrishBoy123 the answer i got from my professor was (1-ye^(xy))/(xe^(xy)+2/y)

Answer 6

.....which is the second last line of my post

the important thing is: do you know how to do it 🤔

Answer 7

if you are learning implicit diff, you need to follow it through that way. so loads of product rule and chain rule.

Answer 8

@IrishBoy123 i understand the implicit part i just don't get how you incorporate chain or product

Answer 9

take it term by term

you are doing this :

\[\dfrac{d}{dx} \left( e^{xy}+\ln y^2 \right) = \dfrac{d}{dx}(x) \]

so start with
\[\dfrac{d}{dx} \left( e^{xy}\right)\]

Answer 10

A solution using the Total Derivative is attached.

Answer 11

The Mathematica program was used to produce the solution above.

Answer 12

so, using the world's favourite letter [u] for calculus substitutions, we have \(\dfrac{d}{dx} \left( e^{xy}\right) = \dfrac{d}{dx} \left( e^{u}\right)\) where \(u = xy \; [= u(x,y)]\)

and \( \dfrac{d}{dx} \left( e^{u}\right) = \dfrac{d}{du} \left( e^{u}\right) .\dfrac{du}{dx} = e^u .\dfrac{du}{dx}\) [chain rule [and fundamental property of e]]

so now we need to find \(\dfrac{du}{dx}\).... and \(\dfrac{du}{dx} =\dfrac{d}{dx}(xy)\)

that means we use the product rule, ie \(\dfrac{d}{dx}(xy) = x.\dfrac{d}{dx}(y) + \dfrac{d}{dx}(x).y\)

there's a lot of work here but this is also a potentially awesome learning experience🍀