`y = log_10(2x) , (5,1)` Find an equation of the tangent line to the graph of the function at the given point

Friday, March 25, 2016

$\displaystyle{y}={\log}_{{10}}{\left({2}{x}\right)},{\left({5},{1}\right)}$ Find an equation of the tangent line to the graph of the function at the given point

$\displaystyle{y}={\log}_{{10}}{\left({2}{x}\right)}$

The line is tangent to the graph of function at (5,1). The equation of the tangent line is _______.

To solve this, we have to determine the slope of the tangent. Take note that the slope of a tangent is equal to the derivative of the function.

To get the derivative of the function, apply the formula $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\log}_{{a}}{\left({u}\right)}\right]}=\frac{{1}}{{{\ln{{\left({a}\right)}}}\cdot{u}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$ .

Applying this, the y' will be:

$\displaystyle{y}'=\frac{{d}}{{\left.{d}{x}}}{\left[{\log}_{{10}}{\left({2}{x}\right)}\right]}$

$\displaystyle{y}'=\frac{{1}}{{{\ln{{\left({10}\right)}}}\cdot{2}{x}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left({2}{x}\right)}$

$\displaystyle{y}'=\frac{{1}}{{{\ln{{\left({10}\right)}}}\cdot{2}{x}}}\cdot{2}$

$\displaystyle{y}'=\frac{{1}}{{{x}{\ln{{\left({10}\right)}}}}}$

Then, plug-in the given point of tangency to get the slope.

$\displaystyle{y}'=\frac{{1}}{{{5}{\ln{{\left({10}\right)}}}}}$

So the line that is tangent to the graph of the function at point (5,1) has a slope of $\displaystyle{m}=\frac{{1}}{{{5}{\ln{{\left({10}\right)}}}}}$ .

To get the equation of the line, apply the point-slope form

$\displaystyle{y}-{y}_{{1}}={m}{\left({x}-{x}_{{1}}\right)}$

Plugging in the values, it becomes:

$\displaystyle{y}-{1}=\frac{{1}}{{{5}{\ln{{\left({10}\right)}}}}}{\left({x}-{5}\right)}$

$\displaystyle{y}-{1}=\frac{{1}}{{{5}{\ln{{\left({10}\right)}}}}}\cdot{x}-\frac{{1}}{{{5}{\ln{{\left({10}\right)}}}}}\cdot{5}$

$\displaystyle{y}-{1}=\frac{{x}}{{{5}{\ln{{\left({10}\right)}}}}}-\frac{{1}}{{{\ln{{\left({10}\right)}}}}}$

$\displaystyle{y}=\frac{{x}}{{{5}{\ln{{\left({10}\right)}}}}}-\frac{{1}}{{{\ln{{\left({10}\right)}}}}}+{1}$

Therefore, the equation of the tangent line is $\displaystyle{y}=\frac{{x}}{{{5}{\ln{{\left({10}\right)}}}}}-\frac{{1}}{{{\ln{{\left({10}\right)}}}}}+{1}$ .

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Friday, March 25, 2016

$\displaystyle{y}={\log}_{{10}}{\left({2}{x}\right)},{\left({5},{1}\right)}$ Find an equation of the tangent line to the graph of the function at the given point

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Friday, March 25, 2016

Find an equation of the tangent line to the graph of the function at the given point

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{y}={\log}_{{10}}{\left({2}{x}\right)},{\left({5},{1}\right)}$ Find an equation of the tangent line to the graph of the function at the given point

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?