# Calculus I

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• #13250

3.10: Use implicit differentiation to find equation of the tangent line to at the point .

#16776

How to solve the Problem:

The point-slope equation, y-y_1=m(x-x_1), has be utilized to find the equation of the tangent line. x_1 and y_1 are given in the description of the problem with the ordered pair (4,1) where 4=x_1 and 1=y_1.

The difficulty is in finding m, the slope of the equation. However, we’re given a function xy-3y^2=1. Taking the derivative of this equation will allow us to find the slope of the function at the point (4,1). Taking the derivative requires implicit differentiation has stated in the problem description.

The steps for implicitly differentiating the equation is as follows:

Step 1: When one implicitly differentiates an equation he takes the derivative of each term in the equation. The terms for the equation of the problem is as follows: xy, -3y^2, and 1. The following steps will show how to differentiate each one of these terms.

Step 2: In order to differentiate the term xy, the product rule must be utilized. The formula for the product is uv’+vu’ where u=x u’=1 v=y v’=y’. The trickiest part is recognizing that when one is implicitly differentiating a function the dependent variable is treating as composite function (with the inner function being the dependent variable itself) and thus the chain rule has to be applied. Since y is the dependent variable the power rule is used in conjunction with the chain rule (with the inner function being y) to arrive at the conclusion that its derivative is y’. That is the derivative of y^1=1*y’=y’. Thus substituting the required values into the product rule formula produces the following: xy’+y*1 which is xy’+y

Step 3: To differentiate the second term, -3y^2, two things must be recognized. Firstly, the -3 is constant so it need not be differentiated and could be left to the side until the differentiation of the term is finished with the result being multiplied by it. Secondly, y^2 must be treated as a composite function with y^2 being the outer term and y itself being the inner term. As a result, the chain rule must be utilized. Differentiating this term would yield -3(2yy’). The -3 is then distributed to yield -6yy’.

Step 4: It is pretty commonsensical that the derivative of the term 1, one the right side of the equation, is simply zero.

Step 5. Now, all the results of the differentiation are put together to produce xy’+y-6yy’=0.

Step 6: The goal is to solve for y’, the derivative of the entire function. This requires y’ to be isolated from the other parts of the equation. The first step is to transpose y on both sides, i.e., subtracting y on both sides. Doing so yields xy’-6yy’=-y. Then, y’ could be factored out from the terms on the left side of the equation. Doing so yields: y'(x-6y)=-y. Finally (x-6y) is transposed on both sides, i.e., it is dividing on both sides. Doing so yields y’=-y/(x-6y).

Step 7: Subsequently, the numerical value of the slope is found by substituting in the values given by the ordered pair (4,1) where x=4 and y=1 into the derivative equation, y’=-y(x-6y). Do so will yield a numerical value of 1/2.

Step 8: m is the slope of the function at the point (4,1) which is 1/2. As mentioned above, x_1=4 and y_1=1. These values are then substituted into the point-slope equation y-y_1=m(x-x_1). Do so will produce the answer for the problem, the equation of the tangent line at the point (4,1), which is y-1=1/2(x-4).

Any feedback, corrections, and questions are welcomed.

-Abdoulaye

#16786

xy-3y^2=1

Use implicit differentiation
(x*1(dy/dx)+1*y) – 6y(dy/dx) = 0

Move all the dy/dx terms to one side
y*1 = 6y(dy/dx) – x*1(dy/dx)

Solve for dy/dx
(dy/dx) = y / 6y – x

Plug in x and y to find the slope
x is 4 y is 1.
dy/dx = 1/ 6-4 = 1/2

Plug in the x and y and the slope into the equation of the tangent line.
Equation formula is ( y-y1=m(x-x1)
which is y – 1 = 1/2(x – 4)

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