Implicit first derivatives and tangent domains
| English | 中文 | Pinyin |
|---|---|---|
| implicit relation | 隐式关系 | yǐn shì guān xì |
A circle has an upper and lower branch. How can we find its tangent gradient without solving for a square root first?
- A circle has an upper and lower branch. How can we find its tangent gradient without solving for a square root first?
- This lesson studies implicit relation 隐式关系: An equation relating x and y without isolating one as an explicit function of the other.
Choose the mathematical structure
- Treat y as a differentiable local function of x. Differentiate the whole relation with respect to x: a function of y brings a dy/dx factor by the chain rule. Collect terms containing dy/dx and solve when its coefficient is nonzero. If the differentiated relation becomes A(x,y)+B(x,y)y′=0, then y′=−A/B where B≠0; the gradient belongs to a chosen local branch.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines implicit relation?
An equation relating x and y without isolating one as an explicit function of the other.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For x²+y²=25, differentiate to 2x+2y y′=0, hence y′=−x/y for y≠0. At (3,4), the tangent gradient is −3/4 and y−4=−(3/4)(x−3); the normal gradient is 4/3. At (5,0) the tangent is vertical, x=5, and no finite dy/dx exists. For x²+xy+y²=7, differentiate the product xy to y+x y′. Then 2x+y+(x+2y)y′=0, so y′=−(2x+y)/(x+2y). At (1,2), which satisfies the relation, the gradient is −4/5. For e^y+x=3, e^y y′+1=0, giving y′=−e^(−y); at (2,0) the gradient is −1.
Implicit first derivatives and tangent domains
Treat y as a differentiable local function of x
Identify the appropriate changing variable and validate the derivative denominator or local branch.
For the circle at (3,4), find dy/dx.
The circle derivative is −x/y; at (3,4) it is −3/4.
Test a tempting shortcut
- Differentiating y² gives 2y y′, not 2y or 2(y′)². The product xy needs both y and x y′. Verify a supplied point lies on the original relation before assigning its tangent. A zero denominator may indicate a vertical tangent or a singular point; the fraction alone cannot classify every case. For example x²+y²=0 has only one isolated real point, not a smooth circle tangent.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Differentiating y² with respect to x gives 2y without a dy/dx factor. This claim is false. Explain which definition or assumption it violates.
For x²+xy+y²=7 at (1,2), find dy/dx.
Collect (x+2y)y′=−(2x+y); the point gives −4/5.
Differentiating y² with respect to x gives 2y without a dy/dx factor.
Differentiating y² gives 2y y′, not 2y or 2(y′)². The product xy needs both y and x y′. Verify a supplied point lies on the original relation before assigning its tangent. A zero denominator may indicate a vertical tangent or a singular point; the fraction alone cannot classify every case. For example x²+y²=0 has only one isolated real point, not a smooth circle tangent.
Interpret a new situation
- Differentiate every term, retain all chain/product factors, collect the y′ terms and state the resulting denominator condition. Evaluate at the original point and use point-slope form for the tangent. A horizontal tangent has zero finite gradient and a vertical normal; a vertical tangent needs its own line equation. This lesson uses first derivatives only, matching G5.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
For e^y+x=3 at (2,0), find dy/dx.
e^y y′=−1; at y=0 this gives −1.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · G. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
An equation relating x and y without isolating one as an explicit function of the other. Choose the relationship, show the method, check its assumptions and interpret the result.