Implicit Differentiation: Fresh Take

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Implicit Differentiation: Fresh Take

Use implicit differentiation to find derivatives and the equations for tangent lines

What is Implicit Differentiation?

The Main Idea

Implicit differentiation is a method to find derivatives of functions that are not explicitly defined in terms of one variable.
Explicit vs. Implicit Functions:
- Explicit: $y = f(x)$ (e.g., $y = x^2 + 1$ )
- Implicit: Relationship between $x$ and $y$ (e.g., $x^2 + y^2 = 25$ )
Treat $y$ as a function of $x$ and use the chain rule when differentiating with respect to $x$ .
General Process:
- Differentiate both sides of the equation with respect to $x$ .
- Group terms with $\frac{dy}{dx}$ on one side.
- Solve for $\frac{dy}{dx}$ .

Find $\frac{dy}{dx}$ for $y$ defined implicitly by the equation $4x^5+ \tan y=y^2+5x$ .

Follow the problem solving strategy, remembering to apply the chain rule to differentiate $\tan y$ and $y^2$ .

$\frac{dy}{dx}=\large \frac{5-20x^4}{\sec^2 y-2y}$

Find $\frac{dy}{dx}$ for the equation $x^3 + y^3 = 6xy$ .

Differentiate both sides with respect to $x$ :

$\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy)$

Apply the chain rule and differentiate:

$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$

Group terms with $\frac{dy}{dx}$ on one side:

$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$

Factor out $\frac{dy}{dx}$ :

$(3y^2 - 6x)\frac{dy}{dx} = 6y - 3x^2$

Solve for $\frac{dy}{dx}$ :

$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$

Finding Tangent Lines Implicitly

The Main Idea

Application of Implicit Differentiation:
- Used to find tangent lines for curves defined by complex equations
Process:
- Find $\frac{dy}{dx}$ using implicit differentiation
- Evaluate $\frac{dy}{dx}$ at the given point to find the slope
- Use point-slope form to find the equation of the tangent line
Advantages:
- Can handle curves not easily expressed as explicit functions
- Simplifies calculations for certain types of curves (e.g., conics)
Key Equations:
- Point-slope form: $y - y_1 = m(x - x_1)$
- Slope-intercept form: $y = mx + b$

Find the equation of the line tangent to the hyperbola $x^2-y^2=16$ at the point $(5,3)$ .

Using implicit differentiation, you should find that $\frac{dy}{dx}=\frac{x}{y}$ .

$y=\frac{5}{3}x-\frac{16}{3}$

Find the equation of the tangent line to the curve $xy + y^2 = 4$ at the point $(1, 1)$ .

Use implicit differentiation to find $\frac{dy}{dx}$ :

$\begin{array}{rcl} \frac{d}{dx}(xy + y^2) &=& \frac{d}{dx}(4) \\ y + x\frac{dy}{dx} + 2y\frac{dy}{dx} &=& 0 \\ \frac{dy}{dx}(x + 2y) &=& -y \\ \frac{dy}{dx} &=& -\frac{y}{x + 2y} \end{array}$

Evaluate $\frac{dy}{dx}$ at $(1, 1)$ :

$\frac{dy}{dx}\big|_{(1,1)} = -\frac{1}{1 + 2(1)} = -\frac{1}{3}$

Use point-slope form to find the equation of the tangent line:

$y - 1 = -\frac{1}{3}(x - 1)$

Convert to slope-intercept form:

$\begin{array}{rcl} y - 1 &=& -\frac{1}{3}x + \frac{1}{3} \\ y &=& -\frac{1}{3}x + \frac{4}{3} \end{array}$

Therefore, the equation of the tangent line is $y = -\frac{1}{3}x + \frac{4}{3}$ .