# How do you find the equation of the tangent line to the graph #y=log_10(2x)# through point (5,1)?

We will need to differentiate the function

yln10 = ln(2x)#

We differentiate the numerator of this expression using the chain rule and the entire function using the quotient rule.

We now find the equation:

For an approximation:

Hopefully this helps!

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To find the equation of the tangent line to the graph (y = \log_{10}(2x)) through point ((5,1)), we first find the derivative of the function. Then, we evaluate the derivative at (x = 5) to find the slope of the tangent line. Finally, we use the point-slope form of a linear equation to write the equation of the tangent line. The derivative of (y = \log_{10}(2x)) is (\frac{1}{x \ln(10)}), and evaluating it at (x = 5) gives us (\frac{1}{5 \ln(10)}). Using the point-slope form with slope (\frac{1}{5 \ln(10)}) and point ((5,1)), the equation of the tangent line is (y - 1 = \frac{1}{5 \ln(10)}(x - 5)).

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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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