caculus (implicit derivates and on)?

Question

um 3 questions plz..

1. A square is inscribed in a circle. How fast is the area of the square changing when the area of the circle is increasing at the rate of 1 inch squared / minute?

2. If tan (x +y ) = x then dy/dx is???

3. suppose f(3)=2
f' (3) =5
f'' (3) = -2

then d^2 / dx^2 of (f^2 (x)) at x=3 is equal to

* then answer to that problem is 42 which made me think that how i solved it was wrong..

Answer 1

1. If the circle has radius r, the diagonal of the square is 2r and so the side length is r√2. So the area of the square is 2r^2 and the area of the circle is πr^2. So if the area of the square is s and the area of the circle is c, we have s = (2/π) c. So ds/dt = (2/π) dc/dt. If dc/dt = 1 in^2 / min then ds/dt = (2/π) in^2/min.

2. Implicitly differentiate both sides w.r.t. x to get
sec^2 (x + y) . (1 + dy/dx) = 1
<=> sec^2 (x+y) dy/dx = 1 - sec^2 (x+y)
<=> dy/dx = cos^2 (x+y) - 1.

3. Let g(x) = f^2 (x). Then
g'(x) = 2f(x) . f'(x)
g''(x) = (2 f(x) . f''(x)) + (2 f'(x) . f'(x))
= 2 f(x) . f''(x) + 2 (f'(x))^2).

So g''(3) = 2(2.(-2)) + 2 (5)^2
= -8 + 50
= 42.