# How do you find the equation of the tangent line to the graph of #f(x)= (ln x)^2# at x=6?

Now slope of tangent is given by slope of curve at that point.

graph{(y-(ln6)^2-ln6/3(x-6))(y-(lnx)^2)=0 [-1.08, 18.92, -1.16, 8.84]}

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To find the equation of the tangent line to the graph of f(x) = (ln x)^2 at x = 6, 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. The derivative of (ln x)^2 is 2(ln x)(1/x) = 2ln x / x.

Next, we substitute x = 6 into the derivative to find the slope of the tangent line at x = 6. The slope is 2ln 6 / 6.

To find a point on the tangent line, we substitute x = 6 into the original function f(x). The point is (6, (ln 6)^2).

Therefore, the equation of the tangent line is y - (ln 6)^2 = (2ln 6 / 6)(x - 6).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- May someone please explain how to complete this problem? (See attached image)

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