Question

Question #f75bd

Answer 1

Let's rewrite this to make our life easier:

`y= 2^(1/2) - 3x^2`

Now take the first derivative:

`2^(1/2)` disappears from the equation.

`x^2 = 2x`. Then multiply by `-3` to give you `-6x`

`dy/dx = -6x`

For the second derivative, derive the first derivative.

`(d^2y)/dx^2 = -6`

The second derivative is negative, meaning that the graph is a maximum.

Answer 2

`-6/sqrt((2-3x^2)^3`

Explanation:

`"assuming "y=sqrt(2-3x^2)`

`"to obtain the first derivative use the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larr"chain rule"`

`y=sqrt(2-3x^2)=(2-3x^2)^(1/2)`

`rArrdy/dx=1/2(2-3x^2)^(-1/2)xxd/dx(2-3x^2)`

`color(white)(rArrdy/dx)=-3x(2-3x^2)^(-1/2)`

`"to obtain the second derivative use the "color(blue)"product rule"`

`"given "f(x)=g(x)h(x)" then"`

`f'(x)=g(x)h'(x)+h(x)g'(x)larr" product rule"`

`g(x)=-3xrArrg'(x)=-3`

`h(x)=(2-3x^2)^(-1/2)rArrh'(x)=3x(2-3x^2)^(-3/2)`

`rArr(d^2y)/(dx^2)=-3x(3x(2-3x^2)^(-3/2))+(2-3x^2)^(-1/2)(-3)`

`color(white)(rArrd^2y/dx^2)=-3(2-3x^2)^(-3/2)[3x^2+2-3x^2]`

`color(white)(rArrd^2y/dx^2)=-6/(sqrt((2-3x^2)^3)`