SOLUTION: Can you help me with this problem?
Solve by means of implicit differentiation:
(e^x) + (e^y) = (e^(x+y)).
Thanks tutors.
Algebra.Com
Question 765977: Can you help me with this problem?
Solve by means of implicit differentiation:
(e^x) + (e^y) = (e^(x+y)).
Thanks tutors.
Found 2 solutions by solver91311, rothauserc:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Just another application of the sum rule and the chain rule.
)
}{dx})
Then second application of the sum and chain rules:
)
And the rest is just a little algebra:
\ =\ e^{x\,+\,y}\ -\ e^x)
}{e^y\left(1\ -\ e^x\right)})
}{1\ -\ e^x})
}{e^x\ -\ 1})
Either of the last two forms are acceptable for an answer.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
Solve by means of implicit differentiation:
(e^x) + (e^y) = (e^(x+y))
==================================================================
recall that the derivative of e^x is e^x and the derivative of e^y is e^y*y'
e^x + e^y*y' = e^(x+y)*derivative of (x+y)
e^x +e^y*y' = e^(x+y)*(1+y')
============================================================
now solve for y'
e^y*y' = e^(x+y) +y'*e^(x+y)
e^y*y' - y'*e^(x+y) = e^(x+y) - e^x
y' = (e^(x+y) - e^x) / (e^y - e^(x+y))
===========================================================
now e^(x+y) = e^x + e^y, so let's substitute
y' = (e^x + e^y - e^x) / (e^y - e^x - e^y)
y' = e^y / -e^x = -(e^(y-x))
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