Derivatives: Solving for t -Xe^(-xt) + Ye^(-yt)

[mrooney]

I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance

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Solving for t

-Xe^-xt + Ye^-yt


Final answer should be
t= ln(y/x)(1/(y-x))

[Subhotosh Khan]

I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance

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Solving for t

-Xe^-xt + Ye^-yt


Final answer should be
t= ln(y/x)(1/(y-x))

That is not an equation - there is no "equal to"(=) sign in there.

[mrooney]

That is not an equation - there is no "equal to"(=) sign in there.

sorry, its supposed to be equal to 0

[Subhotosh Khan]

sorry, its supposed to be equal to 0

So edit your original post.

Also, please explain how are X, x, Y & y are related.

[Harry_the_cat]

I have this equation and I know what the final answer should be but I am stuck on how to get to it using derivatives. Thanks in advance

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Solving for t

-Xe^-xt + Ye^-yt


Final answer should be
t= ln(y/x)(1/(y-x))

Assuming x and X are meant to be the same thing (and that's a BIG assumption) and the same for the Y and y:

[tex]-xe^{-xt} + ye^{-yt} =0 [/tex]

[tex] y*e^{-yt} = x* e^{-xt} [/tex]

[tex]\frac {y}{x} = \frac {e^{-xt}}{e^{-yt}} [/tex]

[tex]\frac{y}{x} = e^{yt-xt}[/tex]

Now take natural log of both sides and see if you can make t the subject.