Question

What is the second derivative of x^2 + y^2 = 25 evaluated at (-3, -4)?

I got 4/3. Is this correct?

Answer 1

It depends on what first derivative you're taking. You can take `d/dx` (which I do below), `dx/dy`or `dy/dx`.

Explanation:

Assuming we want to find the derivative with respect to x, we can treat y as a constant (derivative of a constant is zero).

`d/dx (x^2+y^2=25)`

We can break this up using the sum rule `(a+b)'= a' + b'`

`d/dx (x^2) + d/dx(y^2=25)`

Using the power rule, `d/dx (x^2)` becomes 2x, and if we treat `y^2` as a constant, the derivative of that and 25 becomes 0. We're just left with 2x.

`d/dx=2x`

Finding the Second Derivative:
`d/dx (2x)=2`

Through finding the second derivative, we arrive at 2. Please excuse me if my answer is misleading or incorrect, as I have not delved into calculus for too long.

Answer 2

`(d^2y)/(dx^2) = 25/(64)` at the coordinate `(-3,-4)`

Explanation:

We have:

`x^2 + y^2 = 25`

We should recognise this as a circle of radius `5` centred on the origin.

Differentiating Implicitly wrt `x` we get:

`2x + 2ydy/dx = 0`
`:. ydy/dx = -x`
`:. dy/dx = -x/y`

Nowe we differentiate a second time (implicitly) whilst applying the quotient rule:

`(d^2y)/(dx^2) = - ( (y)(1) - (x)(dy/dx) ) / (y)^2`
`\ \ \ \ \ \ \ = - ( y - (x)(-x/y) ) / (y^2)`
`\ \ \ \ \ \ \ = - ( y + x^2/y) / (y^2)`
`\ \ \ \ \ \ \ = - ( (y^2 + x^2)/y) / (y^2)`
`\ \ \ \ \ \ \ = - ( y^2 + x^2) / (y^3)`
`\ \ \ \ \ \ \ = - 25 / (y^3)`

So at the point `(-3,-4)` we have:

`(d^2y)/(dx^2) = -25/(-4)^3`
`\ \ \ \ \ \ \ = -25/(-64)`
`\ \ \ \ \ \ \ = 25/(64)`