4. Find an equation for the tangent
line to the function
at the point
.
Foundations:
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Since only two variables are present, we are going to differentiate everything with respect to in order to find an expression for the slope, . Then we can use the point-slope equation form at the point to find the equation of the tangent line.
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Note that implicit differentiation will require the product rule and the chain rule. In particular, differentiating can be treated as
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which has as a derivative
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
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Solution:
Finding the slope:
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We use implicit differentiation on our original equation to find
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From here, I would immediately plug in to find :
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, or 
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Writing the Equation of the Tangent Line:
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Now, we simply plug our values of and into the point-slope form to find the tangent line through is
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or in slope-intercept form
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
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