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Implicit Differentiation Example Solution

Created by @nathanedwards
 at November 1st 2023, 3:02:33 pm.
AP_Calculus_AB-Questions
#calculus
#implicit-differentiation
#tangent-line

AP Calculus AB Exam Question - Implicit Differentiation

Consider the implicitly defined equation of a curve:

x3−y3+3xy=6 x^3 - y^3 + 3xy = 6 (a)

(b) Find the equation of the tangent line to the curve at the point (1,1)(1, 1).

Answer

(a) To find dydx\frac{{dy}}{{dx}} using implicit differentiation, we'll differentiate both sides of the equation with respect to xx, treating yy as an implicit function of xx.

Taking the derivative of x3−y3+3xyx^3 - y^3 + 3xy with respect to xx using the chain rule, we get:

ddx(x3−y3+3xy)=ddx(6)\frac{{d}}{{dx}}(x^3 - y^3 + 3xy) = \frac{{d}}{{dx}}(6)

Using the power rule, the chain rule, and the product rule, we can differentiate each term:

3x2−ddx(y3)+3(xdydx+y)=03x^2 - \frac{{d}}{{dx}}(y^3) + 3\left(x\frac{{dy}}{{dx}} + y\right) = 0

We differentiate y3y^3 using the chain rule:

ddx(y3)=3y2⋅dydx\frac{{d}}{{dx}}(y^3) = 3y^2 \cdot \frac{{dy}}{{dx}}

Substituting this back into our equation, we have:

3x2−3y2⋅dydx+3(xdydx+y)=03x^2 - 3y^2 \cdot \frac{{dy}}{{dx}} + 3\left(x\frac{{dy}}{{dx}} + y\right) = 0

Now, let's isolate dydx\frac{{dy}}{{dx}} on one side of the equation:

3x2+3xdydx−3y2dydx+3y=03x^2 + 3x\frac{{dy}}{{dx}} - 3y^2\frac{{dy}}{{dx}} + 3y = 0
3xdydx−3y2dydx=−3x2−3y3x\frac{{dy}}{{dx}} - 3y^2\frac{{dy}}{{dx}} = -3x^2 - 3y
dydx(3x−3y2)=−3x2−3y\frac{{dy}}{{dx}}(3x - 3y^2) = -3x^2 - 3y
dydx=−3x2−3y3x−3y2\frac{{dy}}{{dx}} = \frac{{-3x^2 - 3y}}{{3x - 3y^2}}
dydx=−x2−yx−y2\frac{{dy}}{{dx}} = \frac{{-x^2 - y}}{{x - y^2}}

Hence, the expression for dydx\frac{{dy}}{{dx}} in terms of xx and yy is −x2−yx−y2\frac{{-x^2 - y}}{{x - y^2}}.

(b) To find the equation of the tangent line to the curve at the point (1,1)(1, 1), we'll substitute x=1x = 1 and y=1y = 1 into the expression for dydx\frac{{dy}}{{dx}} derived in part (a).

dydx=−(1)2−(1)(1)−(1)2=−20\frac{{dy}}{{dx}} = \frac{{-(1)^2 - (1)}}{{(1) - (1)^2}} = \frac{{-2}}{{0}}

Here, we encounter an indeterminate form (−20\frac{{-2}}{{0}}). To resolve this, we'll find the limit as xx approaches 11 and yy approaches 11.

Let's differentiate the given equation implicitly to find d2ydx2\frac{{d^2y}}{{dx^2}}:

ddx(3x2+3xdydx−3y2dydx)=ddx(−3x2−3y)\frac{{d}}{{dx}}(3x^2 + 3x\frac{{dy}}{{dx}} - 3y^2\frac{{dy}}{{dx}}) = \frac{{d}}{{dx}}(-3x^2 - 3y)
6x+3(dydx)2+3xd2ydx2−3(dydx)2−6ydydx=−6x−36x + 3\left(\frac{{dy}}{{dx}}\right)^2 + 3x\frac{{d^2y}}{{dx^2}} - 3\left(\frac{{dy}}{{dx}}\right)^2 - 6y\frac{{dy}}{{dx}} = -6x - 3
6x−6ydydx+3xd2ydx2=−6x−36x - 6y\frac{{dy}}{{dx}} + 3x\frac{{d^2y}}{{dx^2}} = -6x - 3
6x−6ydydx+3xd2ydx2=−6x−36x - 6y\frac{{dy}}{{dx}} + 3x\frac{{d^2y}}{{dx^2}} = -6x - 3
6x−6y(−x2−yx−y2)+3xd2ydx2=−6x−36x - 6y\left(\frac{{-x^2 - y}}{{x - y^2}}\right) + 3x\frac{{d^2y}}{{dx^2}} = -6x - 3
6x+6xyx−y2+3xd2ydx2=−6x−36x + \frac{{6xy}}{{x - y^2}} + 3x\frac{{d^2y}}{{dx^2}} = -6x - 3
6xyx−y2+3xd2ydx2=−9x−3\frac{{6xy}}{{x - y^2}} + 3x\frac{{d^2y}}{{dx^2}} = -9x - 3
3x(2yx−y2+d2ydx2)=−9x−33x\left(\frac{{2y}}{{x - y^2}} + \frac{{d^2y}}{{dx^2}}\right) = -9x - 3

Now, substitute x=1x = 1 and y=1y = 1 into this equation to find d2ydx2\frac{{d^2y}}{{dx^2}}:

3(1)(2(1)1−(1)2+d2ydx2)=−9(1)−33(1)\left(\frac{{2(1)}}{{1 - (1)^2}} + \frac{{d^2y}}{{dx^2}}\right) = -9(1) - 3
60+3d2ydx2=−9−3\frac{{6}}{{0}} + 3\frac{{d^2y}}{{dx^2}} = -9 - 3
3d2ydx2=−123\frac{{d^2y}}{{dx^2}} = -12
d2ydx2=−4\frac{{d^2y}}{{dx^2}} = -4

Therefore, the second derivative of yy with respect to xx evaluated at (1,1)(1, 1) is −4-4.

Now, let's use the point-slope form to find the equation of the tangent line:

y−y1=m(x−x1)y - y_1 = m(x - x_1)
y−1=(−4)(x−1)y - 1 = (-4)(x - 1)
y−1=−4x+4y - 1 = -4x + 4
y=−4x+5y = -4x + 5

Hence, the equation of the tangent line to the curve at the point (1,1)(1, 1) is y=−4x+5y = -4x + 5.

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