Final answer:
To find the equation of the tangent line to the curve xy=ex at x=e, we can first find the derivative of the curve. Taking the derivative with respect to x, we get dy/dx = ex = y. Now, we need to find the slope of the tangent line at x=e, which is e^e. Using the point-slope form of the equation of a line, we can find the equation of the tangent line: y-e^e = e^e(x-e) + e^e.
Step-by-step explanation:
To find the equation of the tangent line to the curve xy=ex at x=e, we can first find the derivative of the curve. Taking the derivative with respect to x, we get:
dy/dx = (d/dx)(ex)/y = ex = y.
Now, we need to find the slope of the tangent line at x=e. Substituting x=e and y=e^e into the derivative equation, we have:
dy/dx = (e^e) = e^ex=e^e.
So, the slope of the tangent line at x=e is e^e. Now, we can use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is given by:
y-y1 = m(x-x1),
where m is the slope of the line and (x1, y1) is a point on the line. Substituting m=e^e and (x1, y1) = (e, e^e), we have:
y-e^e = e^e(x-e).
Therefore, the equation of the tangent line to the curve xy=ex at x=e is y = e^e(x-e) + e^e.