Implicit Differentiation - AP Calculus BC

1. Find $\frac{dy}{dx}$ for $\frac{x}{y} = 1$.

   $\frac{dy}{dx} = \frac{y}{x}$. From $\frac{x}{y} = 1$, we get $x = y$, so slopes are equal.

2. Find $\frac{dy}{dx}$ for $x^2 + 2y^2 = 3xy$.

   $\frac{dy}{dx} = \frac{3y - 2x}{4y - 3x}$. Collect $\frac{dy}{dx}$ terms and solve for the derivative.

3. Differentiate $x^3 + y^3 = 6xy$ implicitly.

   $3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$. Apply power rule to each cubic term, product rule to $6xy$.

4. Differentiate $x^2 + y^2 = 1$ implicitly.

   $2x + 2y\frac{dy}{dx} = 0$. Apply $\frac{d}{dx}$ to both sides, using chain rule for $y^2$ term.

5. Differentiate $\text{sin}(xy) = x$ implicitly.

   $\text{cos}(xy)(y + x\frac{dy}{dx}) = 1$. Chain rule: $\frac{d}{dx}[\sin(u)] = \cos(u)\frac{du}{dx}$.

6. What is implicit differentiation?

   Differentiating both sides of an equation with respect to $x$. Used when $y$ is not explicitly solved for in terms of $x$.

7. Differentiate $x^2 + y^2 = 4xy$ implicitly.

   $2x + 2y\frac{dy}{dx} = 4y + 4x\frac{dy}{dx}$. Standard implicit differentiation with product rule on right.

8. Differentiate $\text{cos}(x + y) = x$ implicitly.

   $-\text{sin}(x+y)(1 + \frac{dy}{dx}) = 1$. Chain rule: $\frac{d}{dx}[\cos(u)] = -\sin(u)\frac{du}{dx}$.

9. Find $\frac{dy}{dx}$ for $\text{sin}(xy) = x$.

   $\frac{dy}{dx} = \frac{1 - y\text{cos}(xy)}{x\text{cos}(xy)}$. Solve for $\frac{dy}{dx}$ from the differentiated equation.

10. Find $\frac{dy}{dx}$ for $xy = 1$.

    $\frac{dy}{dx} = -\frac{y}{x}$. Solve for $\frac{dy}{dx}$ from the differentiated equation.

11. Differentiate $\text{sin}(xy) = y$ implicitly.

    $\text{cos}(xy)(y + x\frac{dy}{dx}) = \frac{dy}{dx}$. Chain rule on $\sin(xy)$ with product rule inside.

12. Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$.

    $\frac{dy}{dx} = \frac{2x - y}{x - 2y}$. Collect $\frac{dy}{dx}$ terms and solve for the derivative.

13. Find $\frac{dy}{dx}$ for $\ln(x + y) = x - y$.

    $\frac{dy}{dx} = \frac{x+y-1}{x+y+1}$. Collect $\frac{dy}{dx}$ terms and solve algebraically.

14. Find $\frac{dy}{dx}$ for $e^{x+y} = xy$.

    $\frac{dy}{dx} = \frac{y - e^{x+y}}{e^{x+y} - x}$. Collect $\frac{dy}{dx}$ terms and solve algebraically.

15. Differentiate $x^2y + y^2x = 1$ implicitly.

    $2xy + x^2\frac{dy}{dx} + 2yx\frac{dy}{dx} + y^2 = 0$. Use product rule on both terms: $x^2y$ and $y^2x$.

16. Differentiate $y^3 + 3xy = 6$ implicitly.

    $3y^2\frac{dy}{dx} + 3y + 3x\frac{dy}{dx} = 0$. Power rule on $y^3$, product rule on $3xy$ term.

17. Differentiate $\frac{x}{y} = 1$ implicitly.

    $\frac{y - x\frac{dy}{dx}}{y^2} = 0$. Use quotient rule: $\frac{d}{dx}[\frac{u}{v}] = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

18. Find $\frac{dy}{dx}$ for $x^2y + y^2x = 1$.

    $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2yx}$. Factor out $\frac{dy}{dx}$ from numerator and solve.

19. Find $\frac{dy}{dx}$ for $\text{ln}(x + y) = x - y$.

    $\frac{dy}{dx} = \frac{x+y-1}{x+y+1}$. Collect $\frac{dy}{dx}$ terms and solve algebraically.

20. Differentiate $xy = 1$ implicitly.

    $y + x\frac{dy}{dx} = 0$. Use product rule: $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$.

21. Differentiate $\text{ln}(x + y) = x - y$ implicitly.

    $\frac{1}{x+y}(1 + \frac{dy}{dx}) = 1 - \frac{dy}{dx}$. Chain rule for $\ln(x+y)$ gives $\frac{1}{x+y}(1 + \frac{dy}{dx})$.

22. Find $\frac{dy}{dx}$ for $\frac{1}{x+y} = x-y$.

    $\frac{dy}{dx} = \frac{(x+y)^2 - 1}{1 + (x+y)^2}$. Solve by collecting $\frac{dy}{dx}$ terms on one side.

23. Differentiate $y^2 + yx = 1$ implicitly.

    $2y\frac{dy}{dx} + y + x\frac{dy}{dx} = 0$. Product rule on $yx$ term, power rule on $y^2$ term.

24. Differentiate $x^2 - xy + y^2 = 7$ implicitly.

    $2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. Apply power rule to $x^2$ and $y^2$, product rule to $xy$.

25. Differentiate $e^{xy} = y$ implicitly.

    $e^{xy}(y + x\frac{dy}{dx}) = \frac{dy}{dx}$. Chain rule on left, product rule within exponent.

26. Find $\frac{dy}{dx}$ for $y^2 + yx = 1$.

    $\frac{dy}{dx} = \frac{-y}{x + 2y}$. Factor $\frac{dy}{dx}$ and solve for it algebraically.

27. State the chain rule for differentiation.

    If $y=f(u)$ and $u=g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. Essential for differentiating composite functions.

28. Differentiate $e^{x+y} = xy$ implicitly.

    $e^{x+y}(1+\frac{dy}{dx}) = y + x\frac{dy}{dx}$. Chain rule on left, product rule on right side.

29. Find $\frac{dy}{dx}$ for $x^2 + y^2 = 1$.

    $\frac{dy}{dx} = -\frac{x}{y}$. Solve for $\frac{dy}{dx}$ by isolating it algebraically.

30. Differentiate $\frac{1}{x+y} = x-y$ implicitly.

    $-\frac{1}{(x+y)^2}(1 + \frac{dy}{dx}) = 1 - \frac{dy}{dx}$. Use chain rule on $\frac{1}{x+y} = (x+y)^{-1}$.

31. Differentiate $\text{ln}(xy) = y^2$ implicitly.

    $\frac{1}{xy}(y + x\frac{dy}{dx}) = 2y\frac{dy}{dx}$. Chain rule on $\ln(xy)$ and product rule within.

32. Differentiate $x^2 + 2y^2 = 3xy$ implicitly.

    $2x + 4y\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$. Apply power rule to each term, product rule to $3xy$.

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