How do you find the slope of the line tangent to the graph of #y = ln(x^2)# at the point (4, ln(16))?

Answer 1

The slope of the tangent is #1/2#

First, differentiate using the chain rule:

Let #y = ln(u)# and #u = x^2#. If so, #dy/dx = 1/u xx 2x = (2x)/(x^2)#.
The slope of the tangent is given by substituting your point, #x = a#, into the derivative.
So, #m_"tangent" = (2(4))/(4^2) = 8/16 = 1/2#
Hence, the slope of the tangent is #1/2#.

Practice exercises:

Determine the slopes of the tangents at the indicated points.

a) #y = log_2(4x^3)# at the point #x = 2#.
b) #y = sqrt(3x^2 - 2x - 1)# at the point #x = 5#
c) #y = 1/(2x^3 + x - 2)# at the point #x = 1#

Hopefully this helps, and good luck!

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Answer 2

To find the slope of the line tangent to the graph of y = ln(x^2) at the point (4, ln(16)), we can use the derivative.

First, we find the derivative of y = ln(x^2) using the chain rule. The derivative of ln(u) is 1/u multiplied by the derivative of u. Applying this rule, we get:

dy/dx = (1/(x^2)) * (2x)

Next, we substitute x = 4 into the derivative equation:

dy/dx = (1/(4^2)) * (2 * 4)

Simplifying further:

dy/dx = (1/16) * 8

Finally, we calculate the slope:

dy/dx = 1/2

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Answer from HIX Tutor

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.

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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