Implicit differentiation is one of those topics in calculus that students often find mysterious at first—but once you understand the “why” behind it, it becomes a powerful tool in your mathematical toolkit.
In this post, we’ll break down what implicit differentiation is, why it matters, and how you can master it step by step.
📌 What is Implicit Differentiation?
In most of the differentiation problems you’ve seen, the function is given explicitly—for example, y=x2+3xy = x^2 + 3xy=x2+3x. You can directly differentiate it with respect to xxx.
But sometimes, the relationship between xxx and yyy is not so straightforward. For example: x2+y2=25x^2 + y^2 = 25×2+y2=25
Here, yyy is not isolated, but it’s still a function of xxx. This is called an implicit function, and to differentiate it, we use implicit differentiation.
🔍 Why Do We Need It?
Not all equations can be easily solved for yyy. In geometry, physics, and higher mathematics, implicit relationships are common. For instance, circles, ellipses, and many real-world models are defined implicitly.
To find rates of change or tangents to curves like these, we need to use implicit differentiation.
✍️ How to Do Implicit Differentiation (Step-by-Step)
Let’s take the equation: x2+y2=25x^2 + y^2 = 25×2+y2=25
We’ll differentiate both sides with respect to xxx, remembering that yyy is a function of xxx (even if it’s not written that way).
Step 1: Differentiate both sides
ddx(x2)+ddx(y2)=ddx(25)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)dxd(x2)+dxd(y2)=dxd(25) 2x+2y⋅dydx=02x + 2y \cdot \frac{dy}{dx} = 02x+2y⋅dxdy=0
Note: We used the chain rule for y2y^2y2. Since yyy depends on xxx, differentiating y2y^2y2 gives 2y⋅dydx2y \cdot \frac{dy}{dx}2y⋅dxdy.
Step 2: Solve for dydx\frac{dy}{dx}dxdy
2y⋅dydx=−2x⇒dydx=−xy2y \cdot \frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = \frac{-x}{y}2y⋅dxdy=−2x⇒dxdy=y−x
Boom! 🎉 You just used implicit differentiation to find the slope of the curve at any point on the circle.
💡 Pro Tip: When to Use Chain Rule
Any time you differentiate a term involving yyy (or any function of xxx), apply the chain rule. For example:
- ddx(y3)=3y2⋅dydx\frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx}dxd(y3)=3y2⋅dxdy
- ddx(siny)=cosy⋅dydx\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}dxd(siny)=cosy⋅dxdy
🧠 Practice Problem
Try this on your own: x3+y3=6xyx^3 + y^3 = 6xyx3+y3=6xy
Differentiate both sides with respect to xxx and solve for dydx\frac{dy}{dx}dxdy.
(Feel free to comment your answer below or reach out to me if you want a walkthrough!)
🚀 Summary
- Use implicit differentiation when yyy is not isolated.
- Apply the chain rule when differentiating terms involving yyy.
- Always solve for dydx\frac{dy}{dx}dxdy at the end if you’re asked to find the derivative.
📘 Need More Help?
I offer personalized tutoring for high school and college-level calculus, including a deep dive into topics like implicit differentiation, chain rule, and applications of derivatives.
📅 Book a free consultation to find the right learning path for you!
Happy learning,
Rishabh Kumar
Founder, Mathematics Elevate Academy & Mentor Advanced Mathematics & Statistics
Implicit Differentiation – PDF Viewer
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