The second derivative of the position or distance function is the nineth derivat

Question

The second derivative of the position or distance function is the nineth derivative of f(x) = 18x^9 - 25x^3 + 14x^2 - 8x is if the first derivative of a function is over an interval then the function over the interval find the shape of the tangent line to 3x^2 + y^2 + 2x^2y^2 =8y^2 at the part(1,2) find the critical numbers = f(x) dx^2 - 49 given f(x) = (2x + 5)^4 (x - 4)^1/3.find f(x) and simplity. find all critical numbers a balloon in the form of a sphere is being at the rate of the10 cubic meters for minite find the rate of which the surface area of the sphere is increasing at the instant when the radius of the sphere is 3 meters. Find the first derivative y or dy/dx of the function d(x^2 + xy^3) = 2y^3

Explanation / Answer

a) Accelaration

b) 18*9!

c) decreasing

1) 3x^2 + y^4 + 2x^2y^2 = 8y^2

6x+ 4y^3y' + 2(2xy^2+2yy'x^2) = 16yy'

At (1,2)   6+32y' + 2(8+4y') = 32y'

            6 + 32y'+16+8y' = 32y'

y' = -11/4

2) f(x) = y = Sqrt( x^2-49)

f'(x) = y' = 2x/2Sqrt (x^2 - 49)

For critical point y' = 0 . So x = 0 but x= 0 is not in the domain (Number inside the Sqrt should be grater than zero)

So no critical points.

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