- #1

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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???

1. ln(x+y) = e^3x

(1/x+y)(1+(dy/dx)) = (e^3x)(3)

what can u substititue for y???

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- Thread starter punjabi_monster
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- #1

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1. ln(x+y) = e^3x

(1/x+y)(1+(dy/dx)) = (e^3x)(3)

what can u substititue for y???

Last edited:

- #2

lurflurf

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I take it you are finding dy/dx by implict differentiation?punjabi_monster said: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???

[tex]\log(x+y)=e^{3x}[/tex]

differentiation yeilds

[tex]\frac{1+\frac{dy}{dx}}{x+y}=3e^{3x}[/tex]

solving for dy/dx

[tex]\frac{dy}{dx}=3(x+y)e^{3x}-1[/tex]

observe x+y=exp(exp(3x))

[tex]\frac{dy}{dx}=3e^{e^{3x}}e^{3x}-1[/tex]

simplify

[tex]\frac{dy}{dx}=3e^{3x+e^{3x}}-1[/tex]

observe that one could solve the original equation for y and obtain a more straitforward solution.

[tex]\log(x+y)=e^{3x}[/tex]

solve for y

[tex]y=e^{e^{3x}}-x[/tex]

differentiate

[tex]\frac{dy}{dx}=3e^{e^{3x}}e^{3x}-1[/tex]

- #3

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

- #4

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how do u get e^e^3x

- #5

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By solving the natural log.

- #6

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ok thanks guys.

- #7

HallsofIvy

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