Calculus: Second derivative

[erik05]

Find the second derivative by implicit differentiation:

$$x^3 + y^3 = 6xy$$

Wow, a lengthy question when you have to show all your work. Anyway, for the first derivative I got:

$$\frac {2y-x^2}{y^2-2x}$$

How would one start to calculate the second derivative? I tried the Quotient Rule and got:

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

I'm not too sure what to do now.Would it be easier to bring the denominator to the top to get $$(2y-x^2)(y^2-2x)^-1$$ and then using the chain and product rule? Thanks in advance.

 

[Curious3141]

Don't make the first derivative look like a quotient yet. Keep it in the form of :

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

and differentiate both sides implicitly. Use the product rule and the chain rule.

You'll get the second derivative as an expression in terms of the first derivative, x and y. You need to substitute in the expression you found for the first derivative (the quotient) at this point and simplify. Keep that simplification to the last step. Your final answer should be in terms of x and y only.

EDIT : Although, honestly, even doing it the way you did it is fine. Just take that expression (after the quotient rule) and substitute in the expression you already have for the first derivative and simplify till you get an answer in x and y.

 

[wais]

To find the second derivative, we can use the Quotient Rule again, but it would be easier to simplify the first derivative first. Let's start by rewriting the first derivative as:

\frac {2y-x^2}{y^2-2x} = \frac {2y-x^2}{y^2-2x} \cdot \frac {1}{1} = \frac {2y-x^2}{y^2-2x} \cdot \frac {y^2-2x}{y^2-2x} = \frac {2y^3-2xy^2-2x^2y+2x^3}{(y^2-2x)^2}

Now, we can use the Quotient Rule on this simplified form to find the second derivative. Let's label the first derivative as f(x) and the second derivative as f'(x). Using the Quotient Rule, we get:

f'(x) = \frac {(y^2-2x)(2f(x)-2x) - (2y-x^2)(2yf(x)-2x)}{(y^2-2x)^2}

We can simplify this further by expanding the terms in the numerator and collecting like terms. This will give us the final form of the second derivative:

f'(x) = \frac {2y^3-4xy^2-2x^2y+4x^3}{(y^2-2x)^3}

This process of implicit differentiation can be used to find higher order derivatives as well. It may seem daunting at first, but with practice and understanding of the rules, it becomes easier to grasp. Keep up the good work!

 

[thepatientmental]

To find the second derivative, we can use the quotient rule again, this time with the first derivative as the numerator and the original function as the denominator. So, we have:

\frac{\frac{d}{dx}(\frac{2y-x^2}{y^2-2x})}{\frac{d}{dx}(x^3+y^3-6xy)}

Using the quotient rule, we get:

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

Simplifying this, we get:

\frac{(y^2-2x)(2\frac{d^2y}{dx^2}-2)-(2y-x^2)(2y\frac{d^2y}{dx^2}-2)}{(y^2-2x)^2(3x^2+3y^2-6y\frac{dy}{dx}-6x\frac{dy}{dx}-6y)}

Now, we can use the chain rule and product rule to simplify this further. First, we can rewrite the numerator as:

2\frac{d^2y}{dx^2}(y^2-2x)-2(y^2-2x)-2y(2\frac{dy}{dx})+2x(2y\frac{d^2y}{dx^2}-2)

Then, using the chain rule for the terms involving y, we get:

2\frac{d^2y}{dx^2}(y^2-2x)-2(y^2-2x)-4y\frac{dy}{dx}-4xy\frac{d^2y}{dx^2}+4x

Finally, we can rearrange and simplify to get the second derivative:

\frac{2\frac{d^2y}{dx^2}(y^2-2x)-4xy\frac{d^2y}{dx^2}-2(y^2-2x)-4y\frac{dy}{dx