Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　２．７

　Iｍｐｌｉｃｉｔ　Dｉｆｆｅｒｅｎｔｉａｔｉｏｎ

Ｉｎｔｒｏｄｕｃｔｉｏｎ　ｔｏ　ｉｍｐｌｉｃｉｔ　ｄｉｆｆｅｒｅｎｔｉａｔｉｏｎ

• [0:00]. Introduction to the topic: implicit differentiation.
  • Review of the explicit form of equations 𝐲 = 𝒇(𝒙):
    • 🧩 Example –: 𝐲 = $3𝒙^{2}+4$
    • Definition of explicit form: 𝐲 is isolated on one side of the equation, "𝐲" is given explicitly.
    • Explicit Form: In this form, 𝐲 is expressed directly in terms of 𝒙 (e.g., 𝐲 = $3𝒙^{2}+4$). This makes differentiation straightforward as standard rules can be applied directly.
• [1:30]. Definition of implicit form – [📷image]
  • 🧩 Example –: $𝑦+𝑥𝑦=𝑥$
    • Sometimes we can convert an implicit function into an explicit one.
      • Solve for 𝑦: $𝑦=\dfrac{𝑥}{1+𝑥}$
  • [2:25]. 🧩 Example –: $𝑥^{2}+𝑦^{2}=25$ – [📷image]
    • 𝑦 is not isolated; the variables 𝑥 and 𝑦 are mixed together.
    • Implicit form: Here, 𝑦 is intertwined with 𝑥 in the equation (e.g., $𝑥^{2}+𝑦^{2}=25$). This requires treating 𝑦 as a function of 𝑥 and often necessitates the use of the chain rule during differentiation.
    • [4:18]. Implicit equations can often define more than one function of 𝒙
      • $𝑥^{2}+𝑦^{2}=25 \;\Leftrightarrow\; 𝑦=\pm\sqrt{25-𝑥^{2}}$
        • $𝑦=\sqrt{25-𝑥^{2}}$: Represents the upper semicircle (positive).
        • $𝑦=-\sqrt{25-𝑥^{2}}$: Represents the lower semicircle (negative).
      • Explicit differentiation fails when 𝑦 is not a single function, as in $𝑦=\pm\sqrt{25-𝑥^{2}}$, which represents two semicircles. Implicit differentiation handles this directly by differentiating the entire equation ($𝑥^{2}+𝑦^{2}=25$) without solving for 𝑦, allowing for slopes that depend on both 𝑥 and 𝑦.
        • 'allowing for slopes that depend on both 𝑥 and 𝑦': meaning that the derivative at each point is determined by both the 𝑥 and 𝑦 coordinates, not just by 𝑥 alone, as happens with explicit functions
          • When there are multiple solutions for 𝑦 (for example, $+\sqrt{}$ and $-\sqrt{}$), each point $(𝑥,𝑦)$ can have its own slope, which is calculated using implicit differentiation.

Ｄｅｆｉｎｉｔｉｏｎ　ｏｆ　ｔｈｅ　ｉｍｐｌｉｃｉｔ　ｄｉｆｆｅｒｅｎｔｉａｔｉｏｎ　ｔｅｃｈｎｉｑｕｅ

• [4:55]. Review of the concepts of explicit and implicit forms.
  • Understanding the distinction between explicit and implicit forms is crucial because it determines the method used for differentiation. In explicit differentiation, 𝐲 is expressed directly in terms of 𝒙,
    making it straightforward to apply standard differentiation rules. In contrast, implicit differentiation is necessary when 𝐲 cannot be easily isolated, requiring the application of the chain rule during differentiation.
  • In explicit differentiation, the relationship between 𝐲 and 𝒙 is clear and direct, allowing for straightforward differentiation. However, many real-world problems result in equations where 𝐲 cannot be neatly isolated. Implicit differentiation allows us to find $\dfrac{d𝑦}{d𝒙}$ without rearranging the equation, by treating 𝐲 as a function of 𝒙 and applying differentiation rules accordingly.
• [5:20]. 🧩 Example: $2𝒙^{3}+3𝑦^{3}=9𝑥𝑦$
  • [5:40]. It is not always possible to convert an implicit equation to explicit form.
    • We need another method to find derivatives, instead of just assuming 𝑦 equals a function of 𝑥.
      • There has to be a different way—and there is. It's called implicit differentiation.
        • How to take a derivative for $2𝒙^{3}+3𝑦^{3}=9𝑥𝑦$ without solving for 𝑦.
• [8:02]. 🧩 Example –: $𝒙^{3}+𝐲^{3}=5$ – [📷image]
  • Example of converting implicit equations to explicit forms.
    • Some equations naturally present in implicit form cannot be easily rearranged to solve for 𝐲. Attempting to isolate 𝐲 might be complex or impossible, especially with higher-degree polynomials or transcendental functions. Implicit differentiation provides a systematic way to find derivatives without needing to solve for 𝐲 explicitly.
  • To solve $𝒙^{3}+𝐲^{3}=5$, we could solve for 𝐲, subtract $𝒙^{3}$, and take the cube root, allowing us to find the derivative directly. However, the goal is to learn implicit differentiation, which lets us find the derivative without solving for 𝐲.
    • [9:33]. Key point: You must treat 𝐲 as a function of 𝒙, because if 𝐲 can be expressed in terms of 𝒙, it depends on 𝒙.

Ｓｔｅｐｓ　ｆｏｒ　ｉｍｐｌｉｃｉｔ　ｄｉｆｆｅｒｅｎｔｉａｔｉｏｎ

• [9:50]. ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 1: Differentiate both sides of the equation with respect to 𝑥.
  • from 🧩 Example –: $𝒙^{3}+𝐲^{3}=5$ – [📷image]
  • Remember that you are differentiating with respect to 𝑥, while 𝑦 is a function of 𝑥.
  • You can differentiate term by term if the equation allows.
  • Derivative of both sides:
    • $\dfrac{d}{d𝒙}\big[𝑥^{3}+𝑦^{3}\big]=\dfrac{d}{d𝒙}[5]$
    • $\dfrac{d}{d𝒙}[𝑥^{3}]+\dfrac{d}{d𝒙}[𝑦^{3}]=\dfrac{d}{d𝒙}[5]$
    • $3𝑥^{2}+3𝑦^{2}\cdot\dfrac{d}{d𝒙}[𝑦]=0$
      • When differentiating a term that includes 𝑦, apply the chain rule.
        • This requires treating 𝑦 as a function of 𝒙 and often necessitates the use of the chain rule.
          • 𝑦 is a function of 𝒙 that you do not know explicitly.
        • $\dfrac{d}{d𝒙}[𝑦]$ is the derivative of 𝑦 with respect to 𝒙, represented as $\dfrac{d𝑦}{d𝒙}$.
        • Each time you differentiate 𝑦, you must include a $\dfrac{d𝑦}{d𝒙}$ factor due to the chain rule.
          • It’s just like $\dfrac{d}{d𝒙}(3𝒙^{2}-1)^{4}\;\to\;4(3𝒙^{2}-1)^{3}\cdot\dfrac{d}{d𝒙}[3𝒙^{2}-1]$, but now $3𝒙^{2}-1$ is “𝑦”—an unknown function.
• [19:00]. ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 2: Solve the equation for $\dfrac{d𝑦}{d𝒙}$.
  • $3𝑥^{2}+3𝑦^{2}\cdot\dfrac{d}{d𝒙}[𝑦]=0 \;\;\Rightarrow\;\; \dfrac{d}{d𝒙}[𝑦]=-\dfrac{𝑥^{2}}{𝑦^{2}}$
• [19:58]. ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 3: Convert the implicit derivative to explicit form (optional)
  • Once the implicit derivative $\dfrac{d𝑦}{d𝒙}$ is obtained, the next step is to substitute the solutions for 𝑦 from the original equation.
  • If the original equation has multiple branches (e.g., $𝑦=\pm\sqrt[3]{\,5-𝑥^{3}\,}$), substitute each solution separately to find the specific slope for each branch.
    • For the positive branch $𝑦=\sqrt[3]{\,5-𝑥^{3}\,}$, substitute this expression for 𝑦 in $\dfrac{d𝑦}{d𝒙}$ to get the explicit slope for that branch.
    • For the negative branch $𝑦=-\sqrt[3]{\,5-𝑥^{3}\,}$, do the same, resulting in a different slope for the negative branch.
  • This ensures that you have the correct slope for each portion of the graph, providing a complete understanding of the function's behavior.

Ｅｘａｍｐｌｅｓ　ｏｆ　ｉｍｐｌｉｃｉｔ　ｄｉｆｆｅｒｅｎｔｉａｔｉｏｎ

• [23:30]. 🧩 Example –: $3𝐲^{2}+\sin(𝐲)=4𝒙^{5}$ – [📷image]
  • ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 1: Differentiate both sides with respect to 𝒙
    • $\dfrac{d}{d𝒙}\big[3𝐲^{2}+\sin(𝐲)\big]=\dfrac{d}{d𝒙}[4𝒙^{5}]$
    • $6𝐲\cdot\dfrac{d𝐲}{d𝒙}+\cos(𝐲)\cdot\dfrac{d𝐲}{d𝒙}=20𝒙^{4}$ # Apply the chain rule to both terms with 𝐲
    • $\dfrac{d𝐲}{d𝒙}\cdot\big(6𝐲+\cos(𝐲)\big)=20𝒙^{4}$ # Factor out 𝒅𝐲/𝒅𝒙
  • ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 2: Solve the equation for $\dfrac{d𝑦}{d𝒙}$.
    • $\dfrac{d𝐲}{d𝒙}=\dfrac{20𝒙^{4}}{\,6𝐲+\cos(𝐲)\,}$
  • [28:52]. The key idea is to remember that you always obtain a $\dfrac{d𝑦}{d𝒙}$ when differentiating 𝐲.
  • [30:50]. Implicit differentiation is simply differentiating without being able to solve for 𝐲.
    • implicit differentiation is used when you cannot (or do not) solve for 𝐲 explicitly in terms of 𝒙. Instead, you differentiate both sides of an equation as they are, treating 𝐲 as a function of 𝒙 (i.e., $𝐲=𝐲(𝒙)$) and applying the chain rule where necessary.
• [31:28]. 🧩 Example –: $𝒙\cdot 𝐲=1$ – [📷image]
  • You must use the product rule when differentiating $𝒙\cdot 𝐲$.
  • ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 1: Differentiate both sides of the equation with respect to 𝑥
    • $𝐲+𝒙\cdot\dfrac{d𝑦}{d𝒙}=0$
  • ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 2: Solve the equation for $\dfrac{d𝑦}{d𝒙}$.
    • $\dfrac{d𝑦}{d𝒙}=-\dfrac{𝐲}{𝒙}$
  • ꜱ̲ᴛ̲ᴇ̲ᴩ̲ 3: Convert the implicit derivative to explicit form.
    • $𝒙\cdot 𝐲=1 \;\Rightarrow\; 𝐲=\dfrac{1}{𝒙}$
    • substitute the solutions for 𝑦 from the original equation.
      • $\dfrac{d𝑦}{d𝒙}=-\dfrac{𝐲}{𝒙} \;\Rightarrow\; \dfrac{d𝑦}{d𝒙}=-\dfrac{(1/𝒙)}{𝒙}=-\dfrac{1}{𝒙^{2}}$
  • [36:00]. In implicit differentiation, you must identify when to apply the product rule or the quotient rule.

Ｓｅｃｏｎｄ　ｄｅｒｉｖａｔｉｖｅ　ｏｆ　ａｎ　ｉｍｐｌｉｃｉｔ　ｆｕｎｃｔｉｏｎ

• [36:12]. 🧩 Example –: $3𝒙^{2}-𝐲^{2}=16$ – [📷image]
  • Find the first derivative
    • $\dfrac{d}{d𝒙}[\,3𝒙^{2}-𝐲^{2}\,]=\dfrac{d}{d𝒙}[16]$
    • $6𝒙-2𝐲\cdot\dfrac{d𝐲}{d𝒙}=0$
    • $-2𝐲\cdot\dfrac{d𝐲}{d𝒙}=-6𝒙$
    • $\dfrac{d𝐲}{d𝒙}=\dfrac{3𝒙}{𝐲}$
  • [38:06]. To find the second derivative, differentiate the first derivative with respect to 𝒙.
    • Use the notation $\dfrac{d^{2}𝐲}{d𝒙^{2}}$ for the second derivative.
    • Apply the quotient rule:
      • $\dfrac{d}{d𝒙}\Big[\dfrac{d𝐲}{d𝒙}\Big]=\dfrac{d}{d𝒙}\Big[\dfrac{3𝒙}{𝐲}\Big]$
      • $\dfrac{d^{2}𝐲}{d𝒙^{2}}=\dfrac{\,𝐲\cdot\dfrac{d}{d𝒙}(3𝒙)-3𝒙\cdot\dfrac{d}{d𝒙}(𝐲)\,}{𝐲^{2}}$
      • $\dfrac{d^{2}𝐲}{d𝒙^{2}}=\dfrac{\,𝐲\cdot 3-3𝒙\cdot\dfrac{d𝐲}{d𝒙}\,}{𝐲^{2}}$
    • Substitute the expression found for $\dfrac{d𝐲}{d𝒙}$:
      • $\dfrac{d^{2}𝐲}{d𝒙^{2}}=\dfrac{\,3𝐲-3𝒙\cdot\big(\dfrac{3𝒙}{𝐲}\big)\,}{𝐲^{2}}$
      • $\dfrac{d^{2}𝐲}{d𝒙^{2}}=\dfrac{\,3𝐲-\dfrac{9𝒙^{2}}{𝐲}\,}{𝐲^{2}}$
    • Simplify the result:
      • $\dfrac{d^{2}𝐲}{d𝒙^{2}}=\dfrac{3𝐲^{2}-9𝒙^{2}}{𝐲^{3}}$ # We write $3𝐲$ as $3𝐲=\dfrac{3𝐲^{2}}{𝐲}$ to obtain a common denominator.

Ｇｅｏｍｅｔｒｉｃ　ｉｎｔｅｒｐｒｅｔａｔｉｏｎ　ｏｆ　ｔｈｅ　ｉｍｐｌｉｃｉｔ　ｄｅｒｉｖａｔｉｖｅ

• [44:22]. 🧩 Example – : Find the slopes at $(2,-1)$ and $(2,1)$ for $𝐲^{2}-𝒙+1=0$ – [📷image]
  • Differentiate both sides with respect to 𝒙:
    • $\dfrac{d}{d𝒙}[\,𝐲^{2}-𝒙+1\,]=\dfrac{d}{d𝒙}[0]$ # Differentiate both sides
    • $2𝐲\cdot\dfrac{d𝐲}{d𝒙}-1=0$ # Chain rule for $𝐲^{2}$, derivative of 𝒙 is $1$
    • $2𝐲\cdot\dfrac{d𝐲}{d𝒙}=1$
    • $\dfrac{d𝐲}{d𝒙}=\dfrac{1}{2𝐲}$ # Isolate 𝒅𝐲/𝒅𝒙
  • Evaluate the slope at $(2,-1)$:
    • $\dfrac{d𝐲}{d𝒙}\Big|_{(2,-1)}=\dfrac{1}{2\cdot(-1)}=-\dfrac{1}{2}$ # Substitute $𝐲=-1$
  • Evaluate the slope at $(2,1)$:
    • $\dfrac{d𝐲}{d𝒙}\Big|_{(2,1)}=\dfrac{1}{2\cdot 1}=\dfrac{1}{2}$ # Substitute $𝐲=1$
  • Comment:
    • In implicit differentiation, the slope at a point can depend on both 𝒙 and 𝐲.
    • Here, the derivative formula $\dfrac{d𝐲}{d𝒙}=\dfrac{1}{2𝐲}$ shows that the slope only depends on the 𝐲 value.
    • So, for each point, substitute the corresponding 𝐲 to find the slope.

Ｅｑｕａｔｉｏｎ　ｏｆ　ｔｈｅ　ｔａｎｇｅｎｔ　ｌｉｎｅ　ｔｏ　ａｎ　ｉｍｐｌｉｃｉｔ　ｃｕｒｖｅ

• [47:28]. Implicit differentiation can be used to find the equation of the tangent line to a curve at a given point.
• [48:01]. 🧩 Example –: $4𝒙^{4}+8𝒙^{2}𝐲^{2}-25𝒙^{2}𝐲+16𝐲^{4}=0$ obtain the slope of the tangent line at the point $(2,1)$. – [📷image-1] – [📷image-2]
  • Differentiate both sides with respect to 𝒙:
    • $\dfrac{d}{d𝒙}\big[\,4𝒙^{4}+8𝒙^{2}𝐲^{2}-25𝒙^{2}𝐲+4𝐲^{4}\,\big]=\dfrac{d}{d𝒙}[0]$ # Differentiate term by term
  • Apply the product and chain rules to each term:
    • $16𝒙^{3}+\dfrac{d}{d𝒙}[8𝒙^{2}𝐲^{2}]-\dfrac{d}{d𝒙}[25𝒙^{2}𝐲]+\dfrac{d}{d𝒙}[4𝐲^{4}]=0$
    • $16𝒙^{3}+16𝒙𝐲^{2}+8𝒙^{2}\cdot 2𝐲\cdot\dfrac{d𝐲}{d𝒙}-(25\cdot 2𝒙𝐲+25𝒙^{2}\cdot\dfrac{d𝐲}{d𝒙})+16𝐲^{3}\cdot\dfrac{d𝐲}{d𝒙}=0$
    • $16𝒙^{3}+16𝒙𝐲^{2}\cdot\dfrac{d𝐲}{d𝒙}-50𝒙𝐲-25𝒙^{2}\cdot\dfrac{d𝐲}{d𝒙}+16𝐲^{3}\cdot\dfrac{d𝐲}{d𝒙}=0$ # Collect all terms with 𝒅𝐲/𝒅𝒙
  • Collect all terms with $\dfrac{d𝐲}{d𝒙}$ to one side:
    • $(16𝒙^{2}𝐲-25𝒙^{2}+16𝐲^{3})\cdot\dfrac{d𝐲}{d𝒙}=50𝒙𝐲-16𝒙^{3}-16𝒙𝐲^{2}$ # Move other terms to the right
  • Solve for $\dfrac{d𝐲}{d𝒙}$:
    • $\dfrac{d𝐲}{d𝒙}=\dfrac{\,50𝒙𝐲-16𝒙^{3}-16𝒙𝐲^{2}\,}{\,16𝒙^{2}𝐲-25𝒙^{2}+16𝐲^{3}\,}$ # This is the implicit derivative
    • This result gives the slope of the tangent line at any point $(𝒙,𝐲)$ on the curve.
  • Evaluate $\dfrac{d𝑦}{d𝑥}$ at the point $(2,1)$ to obtain the slope of the tangent line.
    • At the point $(2,1)$:
      • Substitute $𝒙=2$, $𝐲=1$:
      • $\dfrac{d𝐲}{d𝒙}=\dfrac{50\cdot 2\cdot 1-16\cdot 8-16\cdot 2\cdot 1}{16\cdot 4\cdot 1-25\cdot 4+16\cdot 1}$
      • $=\dfrac{100-128-32}{64-100+16}$
      • $=\dfrac{-60}{-20}$
      • $=3$
      • So, $m=3$ # Slope at $(2,1)$
  • Use the point-slope formula to find the equation of the tangent line.
    • Equation of the tangent line at $(2,1)$:
      • Use point-slope form: $𝐲-1=3(𝒙-2)$
      • Simplify: $𝐲-1=3𝒙-6$
        $𝐲=3𝒙-5$ # Final equation of the tangent line at $(2,1)$

Ｒｅｌａｔｅｄ　ｒａｔｅｓ

• [1:07:21]. Introduction to the concept of related rates: word problems involving implicit derivatives.
  • Techniques for solving related rates problems will be presented in the next session.