Show 2nd derivatives with respect to x of both y=(1-x^2)^(1/2) and y=-(1-x^2)^(1/2) are not zero at x=1

Question

for x²+y²=1 show 2nd derivatives with respect to x for both semicircle functions do not exist at x=1 and x=-1.

Answer 1

Ok so I will answer the question for y = (1 -x²)^1/2

1. Solve the first derivative. This will involve chain rule where the inside function = (1 -x²)
2. y' = 1/2(u)^-1/2 * u'
3. plugging u back in, y' = 1/2(1 - x²)^-1/2 * -2x, which can be simplified so, y' = -x(1 - x²)^-1/2
4. Solve for the second derivative. I will keep my negative exponents for now so this will use chain and product rule, where our inside function is (1 - x)², u = -x, and v = (1 -x²)^-1/2 **product rule is uv' +u'v
5. y'' = -x * -1/2(1 - x²)^-3/2 * -2x + -(1 - x²)^-1/2
6. Simplify: y'' = -x²(1 - x²)^-3/2 - (1 - x)^-1/2
7. Now, plug in x = 1 and x = -1 to the second derivative, which yields DNE for both x values.

Repeat this same process for the other function. Also, just noting I didn't make my exponents positive for my final answer because it is difficult to depict on here formatting wise.

Answer 2

x² + y² = 1

y² = 1 - x²

So, y = ± √(1 - x²)

If y = √(1 - x²), then dy/dx = (1/2)(1 - x²)^-1/2(-2x) = -x / √(1 - x²)

dy/dx is undefined at x = 1 and x = -1. So, at x = 1 and x = -1, dy/dx is not differentiable.

Same issue for y = -√(1 - x²).

Note: The graph of x² + y² = 1 is the unit circle (The circle with center (0,0) and radius 1).

When x = 1 or -1, the graph has a vertical tangent (I.e., dy/dx is undefined).