What is the substitution for y in ln(x+y) = e^3x?

[punjabi_monster]

hi there i am having trouble on the following question.

1. ln(x+y) = e^3x
(1/x+y)(1+(dy/dx)) = (e^3x)(3)
what can u substititue for y?

 

[lurflurf]

hi there i am having trouble on 2 quesitons.

1. ln(x+y) = e^3x
(1/x+y)(1+(dy/dx)) = (e^3x)(3)
what can u substititue for y?

I take it you are finding dy/dx by implict differentiation?
\log(x+y)=e^{3x}
differentiation yeilds
\frac{1+\frac{dy}{dx}}{x+y}=3e^{3x}
solving for dy/dx
\frac{dy}{dx}=3(x+y)e^{3x}-1
observe x+y=exp(exp(3x))
\frac{dy}{dx}=3e^{e^{3x}}e^{3x}-1
simplify
\frac{dy}{dx}=3e^{3x+e^{3x}}-1
observe that one could solve the original equation for y and obtain a more straitforward solution.
\log(x+y)=e^{3x}
solve for y
y=e^{e^{3x}}-x
differentiate
\frac{dy}{dx}=3e^{e^{3x}}e^{3x}-1

 

[Nylex]

You're not meant to do people's work for them :/.

 

[punjabi_monster]

how do u get e^e^3x

 

[Knavish]

By solving the natural log.

 

[punjabi_monster]

ok thanks guys.

 

[HallsofIvy]

In general, however, when doing an implicit differentiation, it is not necessary nor desirable to solve for y and then substitute that for y in the expression for the derivative. It is better to leave it in terms of both x and y.