Implicit first derivatives and tangent domains
Введено| English | Значение |
|---|---|
| implicit relation/ɪmˈplɪsɪt rɪˈleɪʃn/ | неявная зависимость |
How can we find its tangent gradient without solving for a square root first?
- A circle has an upper and lower branch.
- 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.
The method
- 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.
For x²+y²=25, why does differentiating y² produce 2y dy/dx rather than 2y?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Implicit differentiation treats y as y(x): d(y²)/dx=2y·y′. Dropping the factor loses the unknown you are solving for; collecting the y′ terms then gives y′=-x/y.
A worked case
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.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The circle derivative is −x/y; at (3,4) it is −3/4.
Where it goes wrong
- 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.
Not true: Differentiating y² with respect to x gives 2y without a dy/dx factor. By the chain rule, the derivative of y² with respect to x is 2y·dy/dx.
For x²+xy+y²=7 at (1,2), find dy/dx.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Collect (x+2y)y′=−(2x+y); the point gives −4/5.
Differentiating y² with respect to x gives 2y without a dy/dx factor.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
By the chain rule, the derivative of y² with respect to x is 2y·dy/dx.
Applying it
- 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.
For e^y+x=3 at (2,0), find dy/dx.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
e^y y′=−1; at y=0 this gives −1.
Match each implicit-differentiation case to its outcome.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Verify the point lies on the relation first; a zero denominator signals a vertical tangent or singular point, not a finite gradient.
In your course
- AQA A-level Mathematics (7357), specification section G.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Implicit relation: An equation relating x and y without isolating one as an explicit function of the other.
- Treat y as a differentiable local function of x.
- Watch for: differentiating y² gives 2y y′, not 2y or 2(y′)².