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

Explain Bot Explain Bot answered
8 months ago

To find the equation of the tangent line, we need to first find the slope of the tangent line at the given point.

Start by finding the derivative of the given curve, y = (3+lnx)^2, with respect to x:

dy/dx = 2(3 + ln(x))*(1/x)

Now, substitute x = 1/e into the derivative equation to find the slope at x = 1/e:

dy/dx = 2(3 + ln(1/e))*(1/(1/e))
= 2(3 + ln(e))*(e)
= 2(3 + 1)*(e)
= 8e

So, the slope of the tangent line at x = 1/e is 8e.

Next, we need to find the y-coordinate of the curve at x = 1/e. Substitute x = 1/e into the original equation:

y(1/e) = (3 + ln(1/e))^2
= (3 + ln(e))^2
= (3 + 1)^2
= 4^2
= 16

Therefore, the y-coordinate of the curve at x = 1/e is 16.

Now that we know the slope (8e) and a point on the line (1/e, 16), we can use the point-slope form of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

Plugging in the values:
y - 16 = 8e(x - 1/e)

Now, we can simplify this equation:

y - 16 = 8ex - 8
y = 8ex - 8 + 16
y = 8ex + 8

Therefore, the equation of the tangent line to the curve y = (3+lnx)^2 at x = 1/e is y = 8ex + 8.

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