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Implicit differentiation of a function of two or more variables

Suppose the function z = f ( x , y ) defines y implicitly as a function y = g ( x ) of x via the equation f ( x , y ) = 0 . Then

d y d x = f / x f / y

provided f y ( x , y ) 0 .

If the equation f ( x , y , z ) = 0 defines z implicitly as a differentiable function of x and y , then

d z d x = f / x f / z and d z d y = f / y f / z

as long as f z ( x , y , z ) 0 .

[link] is a direct consequence of [link] . In particular, if we assume that y is defined implicitly as a function of x via the equation f ( x , y ) = 0 , we can apply the chain rule to find d y / d x :

d d x f ( x , y ) = d d x ( 0 ) f x · d x d x + f y · d y d x = 0 f x + f y · d y d x = 0.

Solving this equation for d y / d x gives [link] . [link] can be derived in a similar fashion.

Let’s now return to the problem that we started before the previous theorem. Using [link] and the function f ( x , y ) = x 2 + 3 y 2 + 4 y 4 , we obtain

f x = 2 x f y = 6 y + 4.

Then [link] gives

d y d x = f / x f / y = 2 x 6 y + 4 = x 3 y + 2 ,

which is the same result obtained by the earlier use of implicit differentiation.

Implicit differentiation by partial derivatives

  1. Calculate d y / d x if y is defined implicitly as a function of x via the equation 3 x 2 2 x y + y 2 + 4 x 6 y 11 = 0 . What is the equation of the tangent line to the graph of this curve at point ( 2 , 1 ) ?
  2. Calculate z / x and z / y , given x 2 e y y z e x = 0 .
  1. Set f ( x , y ) = 3 x 2 2 x y + y 2 + 4 x 6 y 11 = 0 , then calculate f x and f y : f x = 6 x 2 y + 4 f y = −2 x + 2 y 6.
    The derivative is given by
    d y d x = f / x f / y = 6 x 2 y + 4 −2 x + 2 y 6 = 3 x y + 2 x y + 3 .

    The slope of the tangent line at point ( 2 , 1 ) is given by
    d y d x | ( x , y ) = ( 2 , 1 ) = 3 ( 2 ) 1 + 2 2 1 + 3 = 7 4 .

    To find the equation of the tangent line, we use the point-slope form ( [link] ):
    y y 0 = m ( x x 0 ) y 1 = 7 4 ( x 2 ) y = 7 4 x 7 2 + 1 y = 7 4 x 5 2 .

    A rotated ellipse with equation 3x2 – 2xy + y2 + 4x – 6y – 11 = 0 and with tangent at (2, 1). The equation for the tangent is given by y = 7/4 x – 5/2. The ellipse’s major axis is parallel to the tangent line.
    Graph of the rotated ellipse defined by 3 x 2 2 x y + y 2 + 4 x 6 y 11 = 0 .
  2. We have f ( x , y , z ) = x 2 e y y z e x . Therefore,
    f x = 2 x e y y z e x f y = x 2 e y z e x f z = y e x .

    Using [link] ,
    z x = f / x f / y = 2 x e y y z e x y e x = 2 x e y y z e x y e x and z y = f / y f / z = x 2 e y z e x y e x = x 2 e y z e x y e x .
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Find d y / d x if y is defined implicitly as a function of x by the equation x 2 + x y y 2 + 7 x 3 y 26 = 0 . What is the equation of the tangent line to the graph of this curve at point ( 3 , −2 ) ?

d y d x = 2 x + y + 7 2 y x + 3 | ( 3 , −2 ) = 2 ( 3 ) + ( −2 ) + 7 2 ( −2 ) ( 3 ) + 3 = 11 4
Equation of the tangent line: y = 11 4 x + 25 4

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Key concepts

  • The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables.
  • Tree diagrams are useful for deriving formulas for the chain rule for functions of more than one variable, where each independent variable also depends on other variables.

Key equations

  • Chain rule, one independent variable
    d z d t = z x · d x d t + z y · d y d t
  • Chain rule, two independent variables
    d z d u = z x · x u + z y · y u
    d z d v = z x · x v + z y · y v
  • Generalized chain rule
    w t j = w x 1 x 1 t j + w x 2 x 1 t j + + w x m x m t j

For the following exercises, use the information provided to solve the problem.

Let w ( x , y , z ) = x y cos z , where x = t , y = t 2 , and z = arcsin t . Find d w d t .

d w d t = y cos z + x cos z ( 2 t ) x y sin z 1 t 2

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Let w ( t , v ) = e t v where t = r + s and v = r s . Find w r and w s .

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If w = 5 x 2 + 2 y 2 , x = −3 s + t , and y = s 4 t , find w s and w t .

w s = −30 x + 4 y , w t = 10 x 16 y

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If w = x y 2 , x = 5 cos ( 2 t ) , and y = 5 sin ( 2 t ) , find w t .

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If f ( x , y ) = x y , x = r cos θ , and y = r sin θ , find f r and express the answer in terms of r and θ .

f r = r sin ( 2 θ )

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Suppose f ( x , y ) = x + y , u = e x sin y , x = t 2 , and y = π t , where x = r cos θ and y = r sin θ . Find f θ .

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Practice Key Terms 3

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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