Cannot Solve Question -> Help Needed

[AustrianSaurkraut]

I had previously posted about this but as I continued to work on it I got stuck at a place where I cannot even determine the derivative since it does not work out properly.

Here is the question: find dy/dx of e^y=5^x+y

This is what I have completed so far:
ln(e^y) = ln5^(x+y)
y*ln(e) = ln5(x+y)
y = x*ln5 + y*ln5

From here I don't understand how to derive or what exactly to do. I've looked in my notes but this question just continues to stump me. Any help is greatly appreciated.

 

[lex]

Rearrange to get y=...
(Remember ln(5) might look confusing but it is just a constant number. You could make y the subject if y = 3x + 3y or if y = kx + ky).
Then just differentiate.

 

[Steven G]

I would first combine the y's on one side. Then realize that (1-ln(5))and ln(5) is as much as a constant as 17, just like lex pointed out.

 

[AustrianSaurkraut]

Rearrange to get y=...
(Remember ln(5) might look confusing but it is just a constant number. You could make y the subject if y = 3x + 3y or if y = kx + ky).
Then just differentiate.

Firstly, thank you for the tip about it being a constant number that helped! Secondly, after doing the derivative I got

dy/dx = ln5(1) + ln5*(dy/dx)

What am I supposed to do from here? Do I take the dy/dx term to the right?

 

[JeffM]

DO NOT DOUBLE POST.

What did I suggest that you do in the other post when you got to

[MATH]y = x * ln(5) + y * ln(5)[/MATH]
Did you try that?

 

[AustrianSaurkraut]

DO NOT DOUBLE POST.

What did I suggest that you do in the other post when you got to

[MATH]y = x * ln(5) + y * ln(5)[/MATH]
Did you try that?

I tried this but got stuck because I multiplied the terms and ended with y = x * ln(5) + y * ln(5).

From here, I took the derivative:

dy/dx = (1)ln(5) + dy/dx(ln5), it is the dy/dx on this side of the equation that causes me to be confused. Idk what to do next. Also sorry for double posting, won't happen again!

 

[lex]

dy/dx = ln5(1) + ln5*(dy/dx)
What am I supposed to do from here? Do I take the dy/dx term to the right?

Essentially the same as was suggested before - make [MATH]\dfrac{dy}{dx}[/MATH] the subject of the formula. It would have been less confusing to have done it with y, before differentiating, but as I say, it is essentially the same thing.

[MATH]\frac{dy}{dx} = k + k \frac{dy}{dx}[/MATH]
[MATH]\frac{dy}{dx} - k \frac{dy}{dx} =k[/MATH]
Can you continue from here, to make [MATH]\dfrac{dy}{dx}[/MATH] the subject of the formula? (By the way, k is the number ln5).

 

[JeffM]

LOOK BACK AT WHAT I SUGGESTED YOU TRY. I did not suggest immediately differentiating.

[MATH]y = x * ln(5) + y * ln(5) \implies y - y * ln(5) = x * ln(5) \implies y\{1 - ln(5)\} = x * ln(5) \implies y = \dfrac{ln(5)}{1 - ln(5)} * x.[/MATH]
Try differentiating that.

 

[Steven G]

LOOK BACK AT WHAT I SUGGESTED YOU TRY. I did not suggest immediately differentiating.

[MATH]y = x * ln(5) + y * ln(5) \implies y - y * ln(5) = x * ln(5) \implies y\{1 - ln(5)\} = x * ln(5) \implies y = \dfrac{ln(5)}{1 - ln(5)} * x.[/MATH]
Try differentiating that.

As I always say--Calculus is easy! It is the algebra part that can kill you. Differentiating a multiple of x wrt x is trivial and that is the calculus part!