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K12 LibreTexts

5.6: Implicit Differentiation

  • Page ID
    1242
  • The differentiation techniques up to this point have been applied to function definitions (equations) of the form y = f(x). Not all functions can be stated in this form, e.g., when products of x and y are involved. When they cannot, finding the derivative of y may require a different technique.


    Implicit Differentiation

    Consider the equation 2xy=1.

    We want to obtain the derivative dy/dx. One way to do this is to first solve for y, to produce an explicit function of x,

    \[ y=\frac{1}{2x} \nonumber\]

    and then take the derivative on both sides,

    \[ \frac{dy}{dx}=\frac{d}{dx}[\frac{1}{2x}] \nonumber\]

    \[ =\frac{−1}{2x^2} \nonumber\]

    However, there is another way of finding dydx that uses the original equation that is an implicit function of x. We can directly differentiate both sides.

    Find the derivative dy/dx of the equation 2xy=1 without transforming it to an explicit function of x. To do this, directly differentiate both sides:

    \[ \frac{d}{dx}[2xy]=\frac{d}{dx}[1] \nonumber\]

    Using the Product Rule on the left-hand side,

    \[ y\frac{d}{dx}[2x]+2x\frac{d}{dx}[y] = 0 \nonumber\]

    \[ y[2]+2x\frac{dy}{dx}=0 \nonumber\]

    Note that the chain rule is applied when taking the derivative of a term with y and a dydx is included for those terms. Solving for dydx,

    \[ \frac{dy}{dx}=\frac{−2y}{2x+=\frac{−y}{x}} \nonumber\]

    But since \[y=\frac{1}{2x} \nonumber\] substitution gives

    \[ \frac{dy}{dx}=−\frac{1}{x(2x)} \nonumber\]

    \[ =−\frac{1}{2x^2} \nonumber\]

    This result agrees with the previous calculations. This second method of finding a derivative is called implicit differentiation. You may consider the process and say that the first method is easier and faster and there is no reason for the second method. That may be true for this example and some others, but consider the next problem.

    Find dy/dx if 3y2−cosy=x3.

    Good luck finding an explicit function representation of this equation. Let’s try implicit differentiation and see what happens.

    Differentiating both sides of the equation with respect to x and then solving for dy/dx,

    \[ \frac{d}{dx}[3y^2−cosy]=\frac{d}{dx}[x^3] \nonumber\]

    \[ 3\frac{d}{dx}[y^2]−\frac{d}{dx}[cosy]=3x^2 \nonumber\]

    \[ 3(2y\frac{dy}{dx})−(−siny)\frac{dy}{dx}=3x^2 \nonumber\]

    \[ 6y\frac{dy}{dx}+siny\frac{dy}{dx}=3x^2 \nonumber\]

    \[ [6y+siny]\frac{dy}{dx}=3x^2 \nonumber\]

    Solving for dy/dx, we finally obtain

    \[ \frac{dy}{dx}=\frac{3x^2}{6y+siny} \nonumber\]

    In this problem, implicit differentiation provided a workable path to a solution.

    Implicit differentiation can be used to calculate the slope of the tangent line as the problem below shows.

    Find the equation of the tangent line that passes through the point (1, 2) on the graph of 8y3+x2y−x=3.

    The general approach to solving this problem is to:

    1. find dy/dx, then
    2. substitute the point (1, 2) into the derivative to find the slope, and then
    3. use the equation of the line (either the slope-intercept form or the point-intercept form) to find the equation of the tangent line.

    For step 1, finding an explicit function representation of the equation is not obvious. Using implicit differentiation, however, allows differentiation of both sides:

    \[ \frac{d}{dx}[8y^3+x^2y−x]=\frac{d}{dx}[3] \nonumber\]

    \[ 24y^2\frac{dy}{dx}+[(x^2)(1)\frac{dy}{dx}+y(2x)]−1=0 \nonumber\]

    \[ 24y^2\frac{dy}{dx}+x^2\frac{dy}{dx}+2xy−1=0 \nonumber\]

    \[ [24y^2+x^2]\frac{dy}{dx}=1−2xy \nonumber\]

    \[ \frac{dy}{dx}=\frac{1−2xy}{24y^2+x^2} \nonumber\]

    Now, substituting point (1, 2) into the derivative to find the slope,

    \[ \frac{dy}{dx}=\frac{1−2(1)(2)}{24(2)2+(1)^2} \nonumber\]

    \[=\frac{−3}{97} \nonumber\]

    So the slope of the tangent line is −3/97, which is a small value. (What does this tell us about the orientation of the tangent line?)

    Next we need to find the equation of the tangent line. The slope-intercept form is

    \[ y=mx+b \nonumber\]

    where m=−397 and b is the y-intercept. To find b, simply substitute point (1, 2) into the line equation and solve:

    \[ 2=(−397)(1)+b \nonumber\]

    \[ b=\frac{197}{97} \nonumber\]

    Thus the equation of the tangent line is

    \[ y=\frac{−3}{97}x+\frac{197}{97} \nonumber\]

    Note that we could have used the equivalent point-slope form \[ y−y_1=m(x−x_1) \nonumber\]

    To summarize, to find the derivative of an implicit function follow the following steps:

    1. Differentiate both sides with respect to x
    2. Collect all dy/dx terms on the left hand side of the equation and place the other terms without dy/dx on the right hand side.
    3. Factor common dy/dx from all terms
    4. Solve for dy/dx.

    Note that the expression for the derivative dy/dx may involve BOTH x AND y.


    Examples

    Example 1

    Use implicit differentiation to find d2y/dx2 if 5x2−4y2=9.

    \[ \frac{d}{dx}[5x^2−4y^2]= \frac{d}{dx}[9] \nonumber\]

    \[ 10x−8y \frac{dy}{dx}=0 \nonumber\]

    Solving for dy/dx,

    \[ \frac{dy}{dx}= \frac{5x}{4y} \nonumber\]

    Differentiating both sides implicitly again (and using the quotient rule),

    \[ \frac{d^2y}{dx^2}=\frac{(4y)(5)−(5x)(4\frac{dy}{dx})}{(4y)^2} \nonumber\]

    \[ =\frac{20y}{16y^2}−\frac{20x}{16y^2} \frac{dy}{dx} \nonumber\]

    \[ = \frac{5}{4y}−\frac{5x}{4y} \frac{dy}{dx} \nonumber\]

    But since \[ \frac{dy}{dx}=\frac{5x}{4y} \nonumber\]

    we substitute it into the second derivative:

    \[ \frac{d^2y}{dx^2}=\frac{5}{4y}−\frac{5x}{4y}⋅\frac{5x}{4y} \nonumber\]

    \[ \frac{d^2y}{dx^2}=\frac{5}{4y}−\frac{25x^2}{16y^2} \nonumber\]

    This is the second derivative of y.

    Example 2

    Use implicit differentiation to find \[ \frac{d^2y}{dx^2}|_{(x,y)=(2,3)} \nonumber\]

    The next step is to find: \[ \frac{d^2y}{dx^2|_{(x,y)=(2,3)}} \nonumber\]

    \[ \frac{d^2y}{dx^2}|_{(2,3)}=\frac{5}{4(2)}−\frac{25(2)^2}{16(3)^2} \nonumber\]

    \[ =−\frac{5}{72} \nonumber\]

    Example 3

    What does the second derivative represent?

    Since the first derivative of a function represents the rate of change of the function y=f(x) with respect to x, the second derivative represents the rate of change of the function. For example, in kinematics (the study of motion), the speed of an object (y′) signifies the change of position with respect to time but acceleration (y′′) signifies the rate of change of the speed with respect to time.


    Review

    For #1-6, find dy/dx by implicit differentiation.

    1. \[ x^2+y^2=500 \nonumber\]
    2. \[ x^2y+3xy−2=1 \nonumber\]
    3. \[ \frac{1}{x}+ \frac{1}{y}= \frac{1}{2} \nonumber\]
    4. \[ \sqrt{x}−\sqrt{y}= \sqrt{3} \nonumber\]
    5. \[ sin(25xy^2)=x \nonumber\]
    6. \[ tan^3(x^2−y^2)=tan(\frac{π}{4}) \nonumber\])

    For #7-8, use implicit differentiation to find the slope of the tangent line to the given curve at the specified point.

    1. \[ x^2y−y^2x=−1 \nonumber\] at (1,1)
    2. \[ sin(xy)=y \nonumber\] at (π,1)
    3. Find y′′ by implicit differentiation for x3y3=5.
    4. Use implicit differentiation to show that the tangent line to the curve y2=kx at (x0,y0) is given by \[ y_0y= \frac{1}{2}k(x+x_0) \nonumber\] where k is a constant.
    5. Find \[ \frac{d}{dx}(xsin(y)+ysin(x)) \nonumber\]
    6. Find y′ if \[ x^2+xy+y^2=10 \nonumber\]
    7. Find the formula for the line tangent to the curve \[ y^3+2xy^2−x=2 \nonumber\] at the point (1, 1).
    8. Find \[ \frac{d}{dx}(\sqrt{xy}) \nonumber\]
    9. y2+sin(y)=x. Find \[ \frac{d^2y}{dx^2} \nonumber\] in terms of x and y.

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 3.10.


    Vocabulary

    Term Definition
    explicit function An explicit function is a function where it is possible to state the dependent variable as a function only of the independent variable.
    implicit differentiation Implicit differentiation is an application of differentiation on an implicit function that may yield an expression involving both the independent and the dependent variables.
    implicit function An implicit function is a function where the dependent variable is not an explicit function of the independent variable.
    Tangent line A tangent line is a line that "just touches" a curve at a single point and no others.

    Additional Resources

    Video: Implicit Differentiation by Khan Academy

    Practice: Implicit Differentiation

    Real World: The Color Purple