Let's solve these three equations step by step for parts (a), (b), and (c). I will start with equation 1 and proceed sequentially.

Problem 1: $5x^2 - y^3 = 7$

(a) Find $y'$ by implicit differentiation.

Differentiate both sides of the equation with respect to $x$: $\frac{d}{dx}(5x^2) - \frac{d}{dx}(y^3) = \frac{d}{dx}(7).$

This gives: $10x - 3y^2 \frac{dy}{dx} = 0.$

Now, solve for $\frac{dy}{dx}$: $3y^2 \frac{dy}{dx} = 10x \implies \frac{dy}{dx} = \frac{10x}{3y^2}.$

(b) Solve the equation explicitly for $y$ (if possible) and differentiate to get $y'$ in terms of $x$.

Rewrite the equation as: $y^3 = 5x^2 - 7.$

Taking the cube root of both sides: $y = \sqrt[3]{5x^2 - 7}.$

Differentiate this explicitly with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} \left( \sqrt[3]{5x^2 - 7} \right) = \frac{1}{3}(5x^2 - 7)^{-2/3} \cdot 10x = \frac{10x}{3(5x^2 - 7)^{2/3}}.$

(c) Check that the solutions for parts (a) and (b) are consistent.

We need to confirm that: $\frac{10x}{3y^2} = \frac{10x}{3(5x^2 - 7)^{2/3}}.$

Substitute $y = \sqrt[3]{5x^2 - 7}$, which simplifies to $y^2 = (5x^2 - 7)^{2/3}$. This shows that both expressions for $\frac{dy}{dx}$ are indeed consistent.

Problem 2: $6x^4 + y^5 = 2x$

(a) Find $y'$ by implicit differentiation.

Differentiate both sides: $24x^3 + 5y^4 \frac{dy}{dx} = 2.$

Solve for $\frac{dy}{dx}$: $5y^4 \frac{dy}{dx} = 2 - 24x^3 \implies \frac{dy}{dx} = \frac{2 - 24x^3}{5y^4}.$

(b) Solve explicitly for $y$, then differentiate.

Explicitly solving $y^5 = 2x - 6x^4$, we get: $y = \sqrt[5]{2x - 6x^4}.$

Differentiate: $\frac{dy}{dx} = \frac{1}{5}(2x - 6x^4)^{-4/5} \cdot (2 - 24x^3) = \frac{2 - 24x^3}{5(2x - 6x^4)^{4/5}}.$

(c) Verify consistency.

Substitute $y = \sqrt[5]{2x - 6x^4}$ in the implicit differentiation result, confirming consistency.

Problem 3: $\sqrt{x} + \sqrt{y} = 1$

(a) Implicit differentiation.

Differentiate: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0.$

Solve for $\frac{dy}{dx}$: $\frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \implies \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}.$

(b) Solve explicitly for $y$.

Rewrite the equation: $\sqrt{y} = 1 - \sqrt{x}.$

Square both sides: $y = (1 - \sqrt{x})^2 = 1 - 2\sqrt{x} + x.$

Differentiate: $\frac{dy}{dx} = -\frac{1}{\sqrt{x}} + 1.$

(c) Verify consistency.

Substitute $y = (1 - \sqrt{x})^2$ into $-\frac{\sqrt{y}}{\sqrt{x}}$ and check consistency.

Would you like more details or have any questions?

Related Questions:

1. How does implicit differentiation work for higher-order derivatives?
2. Can implicit differentiation be applied to parametric equations?
3. How would the solutions change if the constants in the equations were different?
4. What happens when the equations cannot be solved explicitly for $y$?
5. How does implicit differentiation apply to trigonometric functions?

Tip: Always verify your implicit differentiation result by substituting the original equation and solving explicitly if possible.