Chapter 4: Partial Differentiation
Section 4.3: Chain Rule
Example 4.3.23
If the equation fx,y=0 implicitly defines y=yx, obtain y″x.
Solution
Mathematical Solution
Obtain y′x by differentiating the identity fx,yx≡0 so that fx+fy y′=0⇒y′= −fxfy.
Now differentiate −fx/fy as if it were a function gx,yx. The chain rule gives gx+gy y′, so that
y″
=∂∂x−fxfy+∂∂y−fxfyy′
= −fy fxx−fx fyxfy2−fy fxy−fx fyyfy2−fxfy
= −fxxfy+fx fxyfy2+fx fxyfy2−fx2 fyyfy3
= −fxxfy+2fx fxyfy2−fx2 fyyfy3
Note the use of the quotient rule for differentiation, the replacement of y′ with −fx/fy, and the assumption of equality of fyx and fxy.
Maple Solution - Interactive
Implicit differentiation via the Context Panel System
Write the equation that implicitly defines yx.
Context Panel: Differentiate≻Implicitly Adjust the top portion of the Implicit Differentiation dialog as per Figure 4.3.23(a).
Context Panel: Expand≻Expand
Context Panel: Conversions≻to diff notation (Figure 4.3.23(b)).
Figure 4.3.23(a) Implicit Differentiation dialog
Figure 4.3.23(b) Context Panel option
fx,y=0
f⁡x,y=0
→implicit differentiation
−D1,1⁡f⁡x,y⁢D2⁡f⁡x,y2−2⁢D1,2⁡f⁡x,y⁢D1⁡f⁡x,y⁢D2⁡f⁡x,y+D2,2⁡f⁡x,y⁢D1⁡f⁡x,y2D2⁡f⁡x,y3
= expand
−D1,1⁡f⁡x,yD2⁡f⁡x,y+2⁢D1,2⁡f⁡x,y⁢D1⁡f⁡x,yD2⁡f⁡x,y2−D2,2⁡f⁡x,y⁢D1⁡f⁡x,y2D2⁡f⁡x,y3
→to diff
−∂2∂x2f⁡x,y∂∂yf⁡x,y+2⁢∂2∂x∂yf⁡x,y⁢∂∂xf⁡x,y∂∂yf⁡x,y2−∂2∂y2f⁡x,y⁢∂∂xf⁡x,y2∂∂yf⁡x,y3
Maple Solution - Coded
Initialize
Simplified Maple notation is available if the commands to the right are first executed.
interfacetypesetting=extended:Typesetting:-Suppressfx,y:Typesetting:-Settingsuserep=true:
Apply the implicitdiff command, and temper the result with expand and a convert
convertexpandimplicitdifff,yx,x,x,diff
−fx,xfy+2⁢fx,y⁢fxfy2−fy,y⁢fx2fy3
The result without the conversion of notation
expandimplicitdifff,yx,x,x
The result without the expand and convert operations
implicitdifff,yx,x,x
The best output without the notational advantages of Typesetting
Remove the Typesetting notational improvements.
interfacetypesetting=standard:Typesetting:-Unsuppressfx,y:
convertexpandimplicitdifffx,y,yx,x,x,diff
−∂2∂x2⁢f⁡x,y∂∂y⁢f⁡x,y+2⁢∂2∂y⁢∂x⁢f⁡x,y⁢∂∂x⁢f⁡x,y∂∂y⁢f⁡x,y2−∂2∂y2⁢f⁡x,y⁢∂∂x⁢f⁡x,y2∂∂y⁢f⁡x,y3
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