find an equation of the tangent line to the graph of y = (ln x)^2, at x = 3

y = ln² ( x )

Put x = 3 into the equation y = ln² ( x )

y ( 3 ) = ln² ( 3 )

y´( x ) = dy / dx = 2 ln ( x ) / x

Slope:

m ( x ) = y´( x ) = 2 ln ( x ) / x

Slope of y = ln² ( x ) at x = 3

m = y´( 3 ) = 2 ln ( 3 ) / 3

Equation of the line y = m x + b , for slope m = 2 ln ( 3 ) / 3, passing [ 3 , ln² ( 3 ) ]:

y = m x + b

ln² ( 3 ) = [ 2 ln ( 3 ) / 3 ] ∙ 3 + b

ln² ( 3 ) = 2 ln ( 3 ) + b

ln² ( 3 ) - 2 ln ( 3 ) = b

b = ln² ( 3 ) - 2 ln ( 3 )

y = m x + b

y = [ 2 ln ( 3 ) / 3 ] ∙ x + ln² ( 3 ) - 2 ln ( 3 )

Tangent line to y = ln² ( x ) at x = 3:

y = 2 x ∙ ln ( 3 ) / 3 + ln² ( 3 ) - 2 ln ( 3 )

the 1st derivative is the slope of the tangent line

Of course! Let's find the tangent line to the graph of y = (ln x)^2 at x = 3.

First, we need to find the derivative of the function y = (ln x)^2. Using the chain rule, we get:

dy/dx = 2(ln x)(1/x)
= 2ln x / x

Now, let's find the slope of the tangent line at x = 3. Substitute x = 3 into the derivative and we get:

dy/dx at x = 3 = 2ln 3 / 3

Now that we have the slope, we can use the point-slope formula to find the equation of the tangent line. Since the point of tangency is (3, (ln 3)^2), we have:

y - (ln 3)^2 = (2ln 3 / 3)(x - 3)

And there you have it! The equation of the tangent line to the graph of y = (ln x)^2 at x = 3 is:

y = (2ln 3 / 3)(x - 3) + (ln 3)^2

Now that the math is out of the way, let me tell you a joke to make up for it: Why did the scarecrow win an award? Because he was outstanding in his field!

To find the equation of the tangent line to the graph of y = (ln x)^2 at x = 3, we can use the point-slope form of a linear equation.

Step 1: Find the derivative of the function y = (ln x)^2.
The derivative of y = (ln x)^2 can be found using the chain rule. Let's calculate it.

Let u = ln x
Let f(u) = u^2

By the chain rule, the derivative of y = f(u) = u^2 with respect to x is given by:
dy/dx = f'(u) * du/dx,
where f'(u) is the derivative of f(u) and du/dx is the derivative of u with respect to x.

The derivative of f(u) = u^2 is f'(u) = 2u.
The derivative of u = ln x with respect to x is du/dx = 1/x.

Therefore, dy/dx = f'(u) * du/dx = 2u * (1/x),
dy/dx = 2(ln x) * (1/x),
dy/dx = 2ln x / x.
So, the derivative of y = (ln x)^2 is dy/dx = 2ln x / x.

Step 2: Determine the slope of the tangent line.
At x = 3, we can substitute this value into the derivative equation dy/dx = 2ln x / x.

dy/dx = 2ln(3) / 3.

Therefore, the slope of the tangent line at x = 3 is 2ln(3) / 3.

Step 3: Find the point of tangency.
To find the corresponding y-coordinate (y-value) for x = 3, we substitute x = 3 into the original equation y = (ln x)^2.

y = (ln 3)^2.
Calculate (ln 3)^2 to find the y-coordinate.

Step 4: Write the equation of the tangent line.
Using the point-slope form of a linear equation, y - y1 = m(x - x1), we can write the equation of the tangent line with the slope (m) and the point (x1, y1).

With x1 = 3, y1 = (ln 3)^2, and m = 2ln(3) / 3, the equation of the tangent line is:
y - (ln 3)^2 = (2ln(3) / 3)(x - 3).

This equation represents the equation of the tangent line to the graph of y = (ln x)^2 at x = 3.