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Section 3.6 : Derivatives of Exponential and Logarithm Functions
8. Find the tangent line to \(f\left( x \right) = \ln \left( x \right){\log _2}\left( x \right)\) at \(x = 2\).
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Start SolutionWe know that the derivative of the function will give us the slope of the tangent line so we’ll need the derivative of the function.
\[f'\left( x \right) = \frac{{{{\log }_2}\left( x \right)}}{x} + \frac{{\ln \left( x \right)}}{{x\ln \left( 2 \right)}}\] Show Step 2Now all we need to do is evaluate the function and the derivative at the point in question.
\[f\left( 2 \right) = \ln \left( 2 \right){\log _2}\left( 2 \right) = \ln \left( 2 \right)\hspace{0.25in}\hspace{0.25in}f'\left( 2 \right) = \frac{{{{\log }_2}\left( 2 \right)}}{2} + \frac{{\ln \left( 2 \right)}}{{2\ln \left( 2 \right)}} = 1\] Show Step 3Now all that we need to do is write down the equation of the tangent line.
\[y = f\left( 2 \right) + f'\left( 2 \right)\left( {x - 2} \right) = \ln \left( 2 \right) + \left( 1 \right)\left( {x - 2} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{x - 2 + \ln \left( 2 \right)}}\]