Asked by Jay
What is the equation of the tangent line given the curve:
y = (3+lnx)^2 at x = 1/e
dy/dx = 2[3 + Ln(x))]*1/x
at x = 1/e this is: 4*e
y(1/e) = 4
The equation of the tangent line is thus given by:
4e(x-1/e) + 4
y = (3+lnx)^2 at x = 1/e
dy/dx = 2[3 + Ln(x))]*1/x
at x = 1/e this is: 4*e
y(1/e) = 4
The equation of the tangent line is thus given by:
4e(x-1/e) + 4
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