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Grade 12 · Advanced · 15 minute lesson

Differentiate a relation without solving it into one global function

Use the chain rule when both coordinates vary along a curve.

Lesson 17 of 30 in Grade 12. Take the time you need; the lesson estimate is a guide.

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01 · Read and understand

What you will learn

  • Use the chain rule when both coordinates vary along a curve.
  • Justify the conclusion "At (3,4), the slope is −3/4" using the stated assumptions.

Before you start

Derivative rules and circle equations.

Keep paper nearby. Read the question once for the context, then again to identify what is known and what you need to find.

Start with a question

For x²+y²=25, find dy/dx at (3,4).

Why this math matters

Use the chain rule when both coordinates vary along a curve. This worked micro-lesson connects a precise mathematical condition to a conclusion you can check. The transfer task asks you to change the setting and decide which parts of the reasoning still apply.

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Make a representation of your own.Sketch the quantities or relationships in this question before working through the solution. The cover image sets the learning scene; it does not show this problem’s exact values.

Set up the model

A useful answer starts with clear assumptions:

  • The point lies on the circle.
  • A differentiable local y-as-a-function-of-x branch is used where y≠0.

02 · Work through the example

Follow the reasoning, one step at a time.

Try to predict the next step before reading it. After each calculation, explain why the operation makes sense and how it helps answer the original question.

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See this example unfold.

The complete worked example, one idea at a time.

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Differentiate a relation without solving it into one global function

Paused

Question: Start with the question. Paused.

Question

Start with the question

For x²+y²=25, find dy/dx at (3,4).

Before you calculate

Read what is known and what you need to find. Make a prediction before moving to the first calculation.

Starts paused. Play advances through the full text at a reading pace; pause whenever you need more time. Previous, Next, and the phase buttons let you set your own pace. Playback pauses when this walkthrough leaves the screen or you switch tabs.

Your device’s reduced-motion setting keeps each phase still. Manual controls remain available. The full written solution stays below.

  1. Build the model

    Differentiate to obtain 2x+2y y′=0

    y depends on x along the local branch, so differentiating y² requires y′.

  2. Work through the mathematics

    y′=−x/y when y≠0

    Isolate the derivative while keeping its denominator condition.

  3. Check the conclusion

    At (3,4), the slope is −3/4

    The radius has slope 4/3, and the tangent is perpendicular to it.

The result

At (3,4), the slope is −3/4

The radius has slope 4/3, and the tangent is perpendicular to it.

Common mistakes to catch

  • Do not differentiate y² as 2y without the chain factor.
  • An implicit curve need not be one global function y=f(x).

03 · Practice independently

Try it before revealing the answer.

Use paper or a calculator as needed. Write your units and reasoning, then open the hint or explanation to check your approach.

Practice 1

What happens at (5,0)?

Show a hint

The tangent direction is vertical.

Reveal answer and explanation

dy/dx is not finite there

The formula's zero denominator reflects a genuine failure to express that tangent with a finite slope.

Practice 2

Find the tangent line at (3,4).

Show a hint

Use point-slope form.

Reveal answer and explanation

y−4=−(3/4)(x−3)

It passes through the point with the computed slope.

Take the idea with you

Use an implicit relation to recover local slopes of a constrained motion.

04 · Reflect and continue

Can you explain it in your own words?

Before moving on, explain the main idea without looking at the worked example. Try both practice questions, check your reasoning, and name one mistake you now know how to avoid. Return to a step if you still need support.

Next lesson

Up next: Relate the sensitivity of a function to that of its inverse

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