2.6 Implicit Differentiation

Main ideas / concepts

• Implicit differentiation is needed when you have an equation relating (x) and (y) but you cannot solve for (y) explicitly (i.e., you can’t rewrite the equation in the form (y=f(x))).
• Even when the curve defined by the equation is not a function (it may fail the vertical line test), you can still find slopes of tangent lines by differentiating implicitly.

• The key mechanism is:

  • Differentiate normally with respect to (x) for (x)-terms.
  • Whenever you differentiate a term involving (y) (since (y) depends on (x)), treat (y) as a function and include a factor of (y’) via the chain rule.

• In equations where (x) appears in both the base and the exponent (examples later: (y=x^x), (y=(\cos x)^{\sin x})), you typically log both sides first and then use implicit differentiation.

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Methodology / step-by-step instructions

A) How to do implicit differentiation (treating (y) as a function of (x))

1. Start with an equation like: [ F(x,y)=0 ]

2. Differentiate both sides with respect to (x).

3. Rules during differentiation:

   • Derivatives of (x)-only expressions are computed normally.

   • For any (y) term:

     • Use chain rule:

       • [ \frac{d}{dx}(y^n)=n y^{n-1}y’ ]

       • [ \frac{d}{dx}(k y)=k y’ ]

   • If products occur, use the product rule (as shown in the examples).

4. After differentiating:

   • Collect all terms involving (y’) on one side.
   • Factor out (y’) if possible.
   • Solve for (y’): the instructor notes that (y’) usually won’t be raised to powers (it typically appears multiplied by expressions), making it solvable.

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B) When (x) is in both the base and exponent: use logarithmic differentiation

1. Recognize the form: [ y=(\text{expression involving } x)^{(\text{expression involving } x)} ]

2. Log both sides:

   • Use log rules to transform the exponent into multiplication: [ \ln y = \ln(\text{base})\cdot (\text{exponent}) ]

3. Then differentiate implicitly:

   • Left side: [ \frac{d}{dx}(\ln y)=\frac{y’}{y} ]

   • Right side: apply product rule and chain rule as needed. 4. Multiply by (y) at the end to isolate (y’).

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Example-based lessons (what the video demonstrates)

1) Warm-up chain rule idea: derivative of (y^3)

• If you write the function in terms of (y), then: [ \frac{d}{dx}(y^3)=3y^2y’ ]

• Emphasis: when you encounter (y), differentiate and multiply by (y’) via the chain rule.

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2) Folium of Descartes (implicit differentiation)

• Given equation: [ x^3 + y^3 - 9xy = 0 ]

• Main challenge: cannot solve cleanly for (y) (terms like (y^3) and (y) prevent isolation).

• Differentiation steps shown:

  • [ \frac{d}{dx}(x^3)=3x^2 ]

  • [ \frac{d}{dx}(y^3)=3y^2y’ ]

  • (\frac{d}{dx}(9xy)) uses product rule:

    • [ \frac{d}{dx}(9xy)=9(y + xy’) ]

• Then:

  • Collect (y’) terms
  • Solve for (y’)
• The result is used to find the slope of the tangent line at a specific point (the instructor uses a point corresponding to (x=4), (y=2), as implied by the substitution shown).

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3) Example: find slope for (y^4 = x^2 + y^3) at points

• Differentiation produces:

  • [ \frac{d}{dx}(y^4)=4y^3y’ ]

  • [ \frac{d}{dx}(x^2)=2x ]

  • [ \frac{d}{dx}(y^3)=3y^2y’ ]

• Solve for (y’): [ y’=\frac{2x}{4y^3-3y^2} ]

• Substitutions:

  • At ((0,1)), the slope becomes (0), matching a “horizontal tangent” expectation.
  • At ((0,0)), the slope becomes undefined ((0/0)) because the graph has a sharp corner.
• Key takeaway:
  • Derivatives fail at sharp corners, since left-hand and right-hand slopes differ and the tangent slope is not uniquely defined.

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4) Example: slope for (y^3 = 13x^2) style (implicit product rule usage)

• The video uses an equation of the form (y^3) equals a constant, then differentiates using a product-rule structure in the (x^3) and (y^2) terms.
• Main lesson: use product rule/chain rule as needed, then isolate (y’), then substitute the point ((x,y)=(1,2)) to get the slope.

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5) Derivative of (y=x^x): log both sides, then differentiate implicitly

• Why implicit/log method is needed:
  • Power rule doesn’t apply directly because the exponent isn’t a constant.
  • Exponential rule doesn’t apply directly because the base isn’t a constant.

• Procedure:

  1. Take natural logs: [ \ln y = \ln(x^x)=x\ln x ]

  2. Differentiate: [ \frac{y’}{y}=\ln x + 1 ]

  3. Multiply by (y=x^x): [ y’ = x^x(\ln x + 1) ]

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6) More log-differentiation examples

a) (y = (\cos x)^{\sin x})

• Recognized pattern: (x) in both base and exponent.

• Steps:

  1. [ \ln y = \sin x\cdot \ln(\cos x) ]

  2. Differentiate both sides using product rule + chain rule:

     • (\frac{y’}{y}) becomes an expression involving (\sin x), (\ln(\cos x)), and the derivative of (\ln(\cos x)).
  3. Multiply by (y) to isolate (y’). - The final structure shown is consistent with: - log → product rule → chain rule (subtitles may contain minor algebra typos, but the method remains consistent).

b) (y = (x+1)^{x-1})

• Procedure:

  1. [ \ln y = (x-1)\ln(x+1) ]

  2. Differentiate: [ \frac{y’}{y}=\frac{d}{dx}\big((x-1)\ln(x+1)\big) ]

  3. Use product rule: [ \frac{d}{dx}[(x-1)\ln(x+1)] = 1\cdot\ln(x+1) + (x-1)\cdot\frac{1}{x+1} ]

  4. Multiply by (y=(x+1)^{x-1}): [ y’=(x+1)^{x-1}\left(\ln(x+1)+\frac{x-1}{x+1}\right) ]

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Final review / “how to choose the technique”

• Use implicit differentiation when you cannot solve for (y) explicitly.

• Another common reason to use log/implicit differentiation:

  • When you have a function like [ y=(\text{base involving }x)^{(\text{exponent involving }x)} ] where neither the standard power rule nor the standard exponential rule directly applies.

• Typical outcome after log differentiation:

  • You get a product-rule-heavy expression, often involving a factor of (\frac{y’}{y}) from differentiating (\ln y).

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Speakers / sources featured

• No other speakers or sources are introduced except:
  • Mention of René Descartes (connected to the “folium of Descartes”).
• The video appears to be hosted/instructed by an unnamed instructor (not explicitly identified by name in the subtitles).