Differntiating natural logs of x and y functions

[bob4000]

i have a question I'm attempting as extra work, I have tried the usual method but no luck... find expressionsf for dy/dx in terms of x and y:

ln(x^2+1) + ln(y+1) = x +y

 

[daveb]

Do you know how to do implicit differentiation?

 

[bob4000]

yes, that's the method i tried, but i get the wrong answer

 

[Hootenanny]

If you show your working we'll try to correct you.

 

[bob4000]

dy/dx - 1/ x^2 +1 + 1 / y+1.dy/dx = 1 + dy/dx

=1 /x^2+1 - 1 = dy/dx(1-1/y+1)

dy/dx = (x^2 + 1)^-1 - 1/ 1- (y+1)^-1

 

[Jeff Ford]

Remember $$\frac{d}{dx} \ln(x^2 +1) \neq \frac{1}{x^2+1}$$

You need to use the chain rule to get the proper derivative.

 

[Hootenanny]

Remember $$\frac{d}{dx} \ln(x^2 +1) \neq \frac{1}{x^2+1}$$

You need to use the chain rule to get the proper derivative.

Yes you need to multiply this by the derivative of the bracket