krit.club logo

Calculus - Implicit differentiation (HL)

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Implicit functions are equations where the dependent variable yy is not isolated on one side, represented as f(x,y)=cf(x, y) = c. Examples include circles x2+y2=r2x^2 + y^2 = r^2 or folia x3+y3=3axyx^3 + y^3 = 3axy.

•

Implicit differentiation is used when it is difficult or impossible to solve for yy in terms of xx.

•

The core principle is the Chain Rule: when differentiating a term containing yy with respect to xx, you differentiate with respect to yy and then multiply by dydx\frac{dy}{dx}. For example, ddx(y2)=2ydydx\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}.

•

The Product Rule and Quotient Rule are frequently applied to terms like xyxy or xy\frac{x}{y} during the process.

•

After differentiating all terms, the equation is algebraically rearranged to group all dydx\frac{dy}{dx} terms on one side and factor them out to solve for dydx\frac{dy}{dx}.

•

To find the second derivative d2ydx2\frac{d^2y}{dx^2}, differentiate the first derivative expression implicitly again, and substitute the expression for dydx\frac{dy}{dx} back into the result.

📐Formulae

ddx[f(y)]=f′(y)dydx\frac{d}{dx}[f(y)] = f'(y) \frac{dy}{dx}

ddx[yn]=nyn−1dydx\frac{d}{dx}[y^n] = ny^{n-1} \frac{dy}{dx}

ddx[xy]=xdydx+y\frac{d}{dx}[xy] = x \frac{dy}{dx} + y

ddx[x2y2]=2xy2+2x2ydydx\frac{d}{dx}[x^2y^2] = 2xy^2 + 2x^2y \frac{dy}{dx}

Gradient at (x1,y1)=dydx∣(x1,y1)\text{Gradient at } (x_1, y_1) = \left. \frac{dy}{dx} \right|_{(x_1, y_1)}

💡Examples

Problem 1:

Find the derivative dydx\frac{dy}{dx} for the curve defined by x2+y2−3xy=7x^2 + y^2 - 3xy = 7.

Solution:

  1. Differentiate each term with respect to xx: ddx(x2)+ddx(y2)−ddx(3xy)=ddx(7)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(3xy) = \frac{d}{dx}(7)

  2. Apply differentiation rules: 2x+2ydydx−3(xdydx+y)=02x + 2y \frac{dy}{dx} - 3 \left( x \frac{dy}{dx} + y \right) = 0

  3. Expand and group dydx\frac{dy}{dx} terms: 2x+2ydydx−3xdydx−3y=02x + 2y \frac{dy}{dx} - 3x \frac{dy}{dx} - 3y = 0 (2y−3x)dydx=3y−2x(2y - 3x) \frac{dy}{dx} = 3y - 2x

  4. Solve for dydx\frac{dy}{dx}: dydx=3y−2x2y−3x\frac{dy}{dx} = \frac{3y - 2x}{2y - 3x}

Explanation:

We use the power rule for x2x^2, the chain rule for y2y^2, and the product rule for 3xy3xy. The derivative of a constant (7) is 0. Finally, we rearrange the equation to isolate the derivative.

Problem 2:

Find the equation of the tangent to the curve y3+x2y=10y^3 + x^2y = 10 at the point (3,1)(3, 1).

Solution:

  1. Differentiate implicitly: 3y2dydx+(2xy+x2dydx)=03y^2 \frac{dy}{dx} + \left( 2xy + x^2 \frac{dy}{dx} \right) = 0

  2. Substitute the point (x,y)=(3,1)(x, y) = (3, 1) into the equation: 3(1)2dydx+2(3)(1)+(3)2dydx=03(1)^2 \frac{dy}{dx} + 2(3)(1) + (3)^2 \frac{dy}{dx} = 0 3dydx+6+9dydx=03 \frac{dy}{dx} + 6 + 9 \frac{dy}{dx} = 0 12dydx=−6  ⟹  dydx=−1212 \frac{dy}{dx} = -6 \implies \frac{dy}{dx} = -\frac{1}{2}

  3. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1): y−1=−12(x−3)y - 1 = -\frac{1}{2}(x - 3) 2y−2=−x+32y - 2 = -x + 3 x+2y−5=0x + 2y - 5 = 0

Explanation:

First, we find the gradient by differentiating implicitly. Substituting the coordinates early makes the algebra easier than solving for dydx\frac{dy}{dx} algebraically first. Then we apply the standard line equation formula.