Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
at the point 
| Background Information:
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The equation of the tangent line to at the point is
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where
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Solution:
| Step 1:
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| We use implicit differentiation to find the derivative of the given curve.
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| Using the product and chain rule, we get
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We rearrange the terms and solve for
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| Therefore,
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| and
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| Step 2:
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Therefore, the slope of the tangent line at the point is
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-6(1)-(-2)}{1-4}}\\&&\\&=&\displaystyle {{\frac {4}{3}}.}\end{array}}}
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Hence, the equation of the tangent line to the curve at the point is
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y={\frac {4}{3}}(x-1)-2.}
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| Final Answer:
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y={\frac {4}{3}}(x-1)-2.}
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