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Find y' for `e^(xy)=x^2+y-3,` find equation of line tangent to the graph at (2,0)
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High School Teacher
`e^(xy) = x^2+y - 3`
We will use explicit differentiation to find y'.
==>` (e^(xy))' = (x^2+y-3)'`
==> `(xy)'e^(xy) = 2x + y'`
==>` (xy'+y)e^(xy) = 2x+y'`
==> `xy'e^(xy) + ye^(xy)= 2x+y'`
Now combine terms with y'.
==> `xy'e^(xy) - y' = 2x - ye^(xy)`
==> `y'(xe^(xy) -1) = 2x -ye^(xy)`
==> `y' = (2x-ye^(xy))/(xe^(xy) -1)`
Now will find the tangent line at the point (2,0)
We will use the equation of the line.
==> y-y1 = m (x-x1)
m is the slope which is the derivative y' at the point (2,0).
==> `y'(2,0) = (2*2-0)/(2e^0 -1) = 4/1 = 4`
Then, the slope is m= 4
==> y-0 = 4(x-2)
==> y= 4x -8 is the equation of the tangent line.
Posted by hala718 on June 22, 2012 at 4:01 AM (Answer #1)
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