Question #f75bd

2 Answers
Nov 8, 2017

Let's rewrite this to make our life easier:

y= 2^(1/2) - 3x^2y=2123x2

Now take the first derivative:

2^(1/2)212 disappears from the equation.

x^2 = 2xx2=2x. Then multiply by -33 to give you -6x6x

dy/dx = -6xdydx=6x

For the second derivative, derive the first derivative.

(d^2y)/dx^2 = -6d2ydx2=6

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

Nov 8, 2017

-6/sqrt((2-3x^2)^36(23x2)3

Explanation:

"assuming "y=sqrt(2-3x^2)assuming y=23x2

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

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