Implicit Differentiation

[Krishang]

[math]x^{xy} + y^{exy} = xe^y[/math]
I was given this problem to differentiate, wrt x. I am just unsure of the differentiation of the first part. Do I take it as its own seperate expression and then take logarithm and then do the differentiation? I am unsure about this part. Thank you.

 

[blamocur]

The standard approach for integrating [imath]x^x[/imath] is representing it as [imath]x^x = e^{x \ln x}[/imath]. Do you see how to do the rest?

 

[nasi112]

[math]x^{xy} + y^{exy} = xe^y[/math]
I was given this problem to differentiate, wrt x. I am just unsure of the differentiation of the first part. Do I take it as its own seperate expression and then take logarithm and then do the differentiation? I am unsure about this part. Thank you.

We can't help you if you do not tell us what is e. Is it constant or variable? You also did not say what is the independent variable. Even if wrt x means the independent variable is x, you can't isolate it. So I think your equation is wrong or you made it up.

Though I can help you with this example.
[math]y = x^x[/math][math]\ln y = \ln x^x[/math][math]\ln y = x\ln x[/math][math]\frac{1}{y}y' = \ln x + x\frac{1}{x}[/math][math]\frac{1}{y}y' = \ln x + 1[/math][math]y' = y(\ln x + 1)[/math][math]y' = x^x(\ln x + 1)[/math]

 

[Dr.Peterson]

[math]x^{xy} + y^{exy} = xe^y[/math]
I was given this problem to differentiate, wrt x. I am just unsure of the differentiation of the first part. Do I take it as its own seperate expression and then take logarithm and then do the differentiation? I am unsure about this part. Thank you.

First, you need to give the full instructions for the problem. Presumably you are trying to do implicit differentiation, assuming y is a function of x, and seeking to find dy/dx. Is that correct?

Second, you seem to be expecting to do logarithmic differentiation. You can't do that here, because taking the log of the LHS will not simplify.

Third, as @blamocur implied (referring to differentiation, not integration), you can write [imath]x^{xy}[/imath] as [imath]e^{xy\ln(x)}[/imath] and then use the chain rule and product rule.

We can't help you if you do not tell us what is e. Is it constant or variable?

I presume the intent of this question is to point out that the e in the exponent of [imath]y^{exy}[/imath] is unexpected, so there is some chance that there is a typo. I would expect e to have the same meaning throughout, namely the base of natural logarithms (a constant).

Perhaps an image of the original problem would help us be sure.

 

[Krishang]

I have written the problem correctly. I was given a sheet on which there were these questions. The question looks off with e in the exponent but it was designed like this only.

First, you need to give the full instructions for the problem. Presumably you are trying to do implicit differentiation, assuming y is a function of x, and seeking to find dy/dx. Is that correct?

Second, you seem to be expecting to do logarithmic differentiation. You can't do that here, because taking the log of the LHS will not simplify.

Third, as @blamocur implied (referring to differentiation, not integration), you can write [imath]x^{xy}[/imath] as [imath]e^{xy\ln(x)}[/imath] and then use the chain rule and product rule.


I presume the intent of this question is to point out that the e in the exponent of [imath]y^{exy}[/imath] is unexpected, so there is some chance that there is a typo. I would expect e to have the same meaning throughout, namely the base of natural logarithms (a constant).

Perhaps an image of the original problem would help us be sure.

The logarithm can be used for the first part right? Like the first term, namely [math]r = x^{xy}[/math][math]\ln(r) = xy \ln(x)[/math]Then, [math]\frac{r'}{r} = y + (xy'+y)\ln(x)[/math]
Then I can transfer r to the other place.
I think this is how to do right for the first term in my question?

 

[Dr.Peterson]

I have written the problem correctly. I was given a sheet on which there were these questions. The question looks off with e in the exponent but it was designed like this only.

Okay.

The logarithm can be used for the first part right? Like the first term, namely [math]r = x^{xy}[/math][math]\ln(r) = xy \ln(x)[/math]Then, [math]\frac{r'}{r} = y + (xy'+y)\ln(x)[/math]
Then I can transfer r to the other place.
I think this is how to do right for the first term in my question?

Yes, you can do it that way, though that is not the only way. I would normally do this (but likely for all terms at once):

[math]\frac{d}{dx}\left(x^{xy}\right)=\frac{d}{dx}\left(e^{xy\ln x}\right)=e^{xy\ln x}\frac{d}{dx}\left(xy\ln x\right)=x^{xy}\left(y\ln x+xy'\ln x+xy\frac{1}{x}\right)=\dots[/math]which gives the same result.

Now continue (either way).

 

[Krishang]

Okay.

Yes, you can do it that way, though that is not the only way. I would normally do this (but likely for all terms at once):

[math]\frac{d}{dx}\left(x^{xy}\right)=\frac{d}{dx}\left(e^{xy\ln x}\right)=e^{xy\ln x}\frac{d}{dx}\left(xy\ln x\right)=x^{xy}\left(y\ln x+xy'\ln x+xy\frac{1}{x}\right)=\dots[/math]which gives the same result.

Now continue (either way).

Your method is faster. But I had a query. We did basically
[math]u = xy \ln(x)[/math][math]\frac{dr}{dx} = \frac {dr}{du} \frac{du}{dx}[/math]
Making the first one [imath]e^u[/imath] basically and we did like that, right? I have a bit of problem right now with the chain rule so I usually understand this way haha.