Asked by Anonymous
Use the process of implicit differentiation to find
dy/dx given that x^3e^y-ye^x=0
dy/dx given that x^3e^y-ye^x=0
Answers
Answered by
oobleck
x^3e^y-ye^x=0
Using the product and chain rules,
3x^2 e^y + x^3 e^y y' - e^x y - e^x y' = 0
Now just solve for y'.
Using the product and chain rules,
3x^2 e^y + x^3 e^y y' - e^x y - e^x y' = 0
Now just solve for y'.
Answered by
Anonymous
Ok. How to solve it that's my issue?
Answered by
oobleck
oh come on. The calculus is done. The rest is just algebra I.
(x^3 e^y - e^x) y' = e^x y - 3x^2 e^y
Now just divide to get y'.
(x^3 e^y - e^x) y' = e^x y - 3x^2 e^y
Now just divide to get y'.
Answered by
oooo
Divide by 0?
Answered by
oobleck
huh? HUH?
y' = (e^x y - 3x^2 e^y) / (x^3 e^y - e^x)
Looks like you need to review implicit differentiation. In general, y' will be an expression involving both x and y.
y' = (e^x y - 3x^2 e^y) / (x^3 e^y - e^x)
Looks like you need to review implicit differentiation. In general, y' will be an expression involving both x and y.
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