Implicit Differentiation: Finding Derivatives of Implicit Functions

1. Why We Need Implicit Differentiation

Let's start with the "why." Why can't we just use our existing rules? And what does it even mean to find a derivative when $y$ isn't a straightforward function of $x$? The following video provides a brilliant intuition for this.

The key insight is that when we have an implicit equation like $x^2+y^2=25$, we are interested in points $(x,y)$ that satisfy this condition. If we move a tiny amount along the curve, the values of $x$ and $y$ change, but they must change in such a way that the equation remains true. Implicit differentiation is the tool that captures this relationship between their rates of change.

2. The Mechanics of Implicit Differentiation

The intuitive idea of dx and dy from the video is formalised by applying our existing differentiation rules, with one crucial addition based on the Chain Rule.

The core principle is: Differentiate both sides of the equation with respect to $x$.

• Terms with only $x$: Differentiate as usual (e.g., $\frac{d}{dx}(x^3)=3x^2$).
• Constant terms: Differentiate to zero (e.g., $\frac{d}{dx}(25)=0$).
• Terms with $y$: This is the key step. Since we assume $y$ is some function of $x$, we must use the Chain Rule.

To understand this, let's briefly revisit the chain rule.

As the text explains, the rule is:
$\frac{d}{dx}[f(y)]=\frac{d}{dy}[f(y)]\times \frac{dy}{dx}=f^{\prime }(y)\frac{dy}{dx}$

In simple terms: you differentiate the $y$ expression as if it were the variable, and then you multiply the result by $\frac{dy}{dx}$.

• Example: $\frac{d}{dx}(y^3)=3y^2\cdot \frac{dy}{dx}$
• Example: $\frac{d}{dx}(\sin y)=\cos y\cdot \frac{dy}{dx}$

The Step-by-Step Process

The entire process can be summarised in a few steps.

Here are the steps written out:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply all necessary rules (Power, Product, Quotient, etc.).
3. For every term involving $y$, apply the Chain Rule: differentiate with respect to $y$ and multiply by $\frac{dy}{dx}$.
4. Get all terms with $\frac{dy}{dx}$ on one side of the equation and all other terms on the other side.
5. Factorise (factor) out $\frac{dy}{dx}$.
6. Solve for $\frac{dy}{dx}$ by dividing.

3. Worked Example

Let's see this process in action. The following video provides a clear, step-by-step walkthrough of a typical problem, making sure to explain the role of the Chain Rule.

Let's work through one more example that involves the Product Rule. Find $\frac{dy}{dx}$ for the equation $x^{2}y+\sin (x)=y^3$.

1. Differentiate both sides with respect to $x$:
   $\frac{d}{dx}(x^{2}y)+\frac{d}{dx}(\sin x)=\frac{d}{dx}(y^3)$

2. Apply the rules:

   • The term $x^{2}y$ is a product, so we use the product rule: $\frac{d}{dx}(uv)=u{v}^{\prime }+v{u}^{\prime }$. Here $u=x^2$ and $v=y$.
     $\frac{d}{dx}(x^{2}y)=x^2\cdot \frac{d}{dx}(y)+y\cdot \frac{d}{dx}(x^2)=x^2\frac{dy}{dx}+y(2x)$
   • The term $\sin (x)$ is standard: $\frac{d}{dx}(\sin x)=\cos x$.
   • The term $y^3$ requires the chain rule: $\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}$.

3. Combine everything:
   $x^2\frac{dy}{dx}+2xy+\cos x=3y^2\frac{dy}{dx}$

4. Isolate $\frac{dy}{dx}$ terms:
   $2xy+\cos x=3y^2\frac{dy}{dx}-x^2\frac{dy}{dx}$

5. Factorise out $\frac{dy}{dx}$:
   $2xy+\cos x=(3y^2-x^2)\frac{dy}{dx}$

6. Solve for $\frac{dy}{dx}$:
   $\frac{dy}{dx}=\frac{2xy+\cos x}{3y^2-x^2}$

Notice that the final expression for the gradient depends on both $x$ and $y$. This is typical for implicit differentiation. To find the numerical gradient at a specific point, you need both the x- and y-coordinates of that point.

4. An Elegant Application: Deriving Other Derivative Rules

Implicit differentiation is not just for oddly shaped curves. It's a powerful theoretical tool that can be used to find the derivatives of inverse functions. In the last lesson, you learned that the derivative of $\ln (x)$ is $\frac{1}{x}$. We can prove this very elegantly using implicit differentiation.

Let's see how. We start with $y=\ln (x)$.

This method is a fantastic example of the interconnectedness of mathematical concepts. The same technique is used to find the derivatives of inverse trigonometric functions like $\arcsin (x)$ and $\arctan (x)$.

Conclusion

You have now added another major technique to your differentiation toolkit. Implicit differentiation frees you from the constraint of only working with explicit functions of the form $y=f(x)$.

Key Takeaways:

• Implicit functions are equations where $y$ is not explicitly isolated.
• To differentiate implicitly, you differentiate both sides with respect to $x$.
• The crucial step is applying the Chain Rule to any term containing $y$: $\frac{d}{dx}[f(y)]=f^{\prime }(y)\frac{dy}{dx}$.
• After differentiating, you rearrange the equation algebraically to solve for $\frac{dy}{dx}$.
• The resulting expression for the gradient often depends on both $x$ and $y$.

In our next lesson, we will begin exploring the many applications of the derivative. Our first application will be a natural follow-on from this topic: using the derivative $\frac{dy}{dx}$ to find the equation of the tangent line to a curve at a specific point.