What is the equation of the tangent line of #f(x)=ln(3x)^2-x^2# at #x=3#?

Question

Emily Ammerman · Accepted Answer

# y = (2/3ln9-6)x +ln^2 9 - 2ln9+9 #

We have # f(x) = ln^2(3x)-x^2 #

The gradient of the tangent at any particular point is given by the derivative at that point.

Differentiating wrt #x# (using the chain rule);

# \ \ \ \ \ f'(x) = 2ln(3x)*1/(3x)*3-2x #
# :. f'(x) = (2ln(3x))/x - 2x #

We need to find #f(x)# and #f'(x)# when #x=3#

#x=3 => f(3) \ \ = ln^2 9-9#
#x=3 => f'(3) = 2/3ln9-6#

So the tangent we seek passes through the point #(3, ln^2 9-9)# and has slope #m=2/3ln9-6#, so using #y-y_1=m(x-x_1)# the required equation is;

# y - (ln^2 9-9) = (2/3ln9-6)(x - 3) #
# :. y - (ln^2 9-9) = (2/3ln9-6)(x - 3) #
# :. y - ln^2 9 + 9 = (2/3ln9-6)x - 3(2/3ln9-6) #
# :. y - ln^2 9 + 9 = (2/3ln9-6)x - 2ln9+18 #
# :. y = (2/3ln9-6)x +ln^2 9 - 2ln9+9 #

We can confirm this solution is correct by looking at the graph:

Charles Anctil · Answer

undefined To find the equation of the tangent line of f(x)=ln(3x)^2-x^2 at x=3, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the chain rule and power rule:

f'(x) = 2(ln(3x))(1/3x)(3) - 2x

Next, we substitute x=3 into f'(x) to find the slope of the tangent line:

f'(3) = 2(ln(3(3)))(1/3(3))(3) - 2(3)

Finally, we evaluate f'(3) to find the slope:

f'(3) = 2(ln(9))(1/9)(3) - 6

Therefore, the slope of the tangent line at x=3 is f'(3).

To find a point on the line, we substitute x=3 into f(x):

f(3) = ln(3(3))^2 - 3^2

Finally, we evaluate f(3) to find the y-coordinate of the point:

f(3) = ln(9) - 9

Therefore, a point on the tangent line is (3, ln(9) - 9).

Using the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept, we can write the equation of the tangent line:

y = f'(3)(x - 3) + f(3)

Substituting the values we found earlier, the equation of the tangent line is:

y = (2(ln(9))(1/9)(3) - 6)(x - 3) + (ln(9) - 9)