链式法则后续学习步骤及(5x-y)^4+2y^3=1294的隐函数求导咨询

Hey there! Let's break down your two questions one by one, with clear, practical guidance:

1. What to learn after mastering the chain rule? #

The chain rule is a foundational tool—here are the most logical next steps to build on it:

• Implicit differentiation: This is exactly what your second question focuses on! It uses the chain rule to find derivatives of functions where $y$ isn't explicitly solved for $x$ (like the equation you're working with). It's essential for analyzing curves that don't pass the vertical line test.
• Combining chain rule with product/quotient rules: Most real-world functions are composites and products/quotients. For example, finding the derivative of $e^{x2} \cdot \sin(3x)$ requires both the product rule and chain rule. Mastering this combination lets you tackle nearly any single-variable derivative problem.
• Higher-order derivatives: Once you can compute first derivatives, move on to second, third, or nth derivatives. You'll use the chain rule repeatedly here, and it's key for topics like concavity, inflection points, and acceleration in physics.
• Related rates problems: This is a fun, practical application of the chain rule. You'll relate the rates of change of interconnected quantities (e.g., how fast water drains from a cone, or how fast a ladder slides down a wall). It's great for seeing calculus in action.
• Partial derivatives (if moving to multivariable calculus): If you're advancing beyond single-variable math, the chain rule extends to functions with multiple variables. This is critical for optimization, vector calculus, and fields like engineering or economics.

2. Fixing your implicit differentiation mistake #

Let's start by pointing out the two key errors in your attempt:

1. The right-hand side of your equation is a constant (1294), so its derivative with respect to $x$ is 0, not 1294. Constants always differentiate to zero!
2. When applying the chain rule to $(5x - y)^3$, you forgot that $y$ is a function of $x$. That means the derivative of $-y$ with respect to $x$ is $-\frac{dy}{dx}$ (often written as $-y'$ for shorthand)—you only multiplied by 5, but missed this critical term.

Now let's redo the differentiation step-by-step for the equation:
(5x - y)^4 + 2y^3 = 1294

1. Differentiate both sides with respect to $x$:

   • First term: Use the chain rule: $4(5x - y)^3 \cdot \frac{d}{dx}(5x - y)$. The derivative of $5x$ is 5, and the derivative of $-y$ is $-y'$, so this becomes $4(5x - y)^3(5 - y')$.
   • Second term: Chain rule again applies to $y^3$ (since $y$ depends on $x$): $2 \cdot 3y^2 \cdot y' = 6y^2 y'$.
   • Right-hand side: $\frac{d}{dx}(1294) = 0$.

2. Combine all terms:

   4(5x - y)^3(5 - y') + 6y^2 y' = 0
   

3. Expand the first term:

   20(5x - y)^3 - 4(5x - y)^3 y' + 6y^2 y' = 0
   

4. Collect all terms with $y'$ on one side, and the rest on the other:

   -4(5x - y)^3 y' + 6y^2 y' = -20(5x - y)^3
   

5. Factor out $y'$:

   y' \left[ -4(5x - y)^3 + 6y^2 \right] = -20(5x - y)^3
   

6. Multiply both sides by -1 to simplify signs:

   y' \left[ 4(5x - y)^3 - 6y^2 \right] = 20(5x - y)^3
   

7. Solve for $y'$ (which is $\frac{dy}{dx}$):

   y' = \frac{20(5x - y)^3}{4(5x - y)^3 - 6y^2}
   

8. Simplify by dividing numerator and denominator by 2:

   y' = \frac{10(5x - y)^3}{2(5x - y)^3 - 3y^2}
   

If you have a specific $(x,y)$ point that lies on the original curve, you can plug those values into this expression to get a numerical derivative at that point.

内容的提问来源于stack exchange，提问作者Tinsley Maryuri