Question #f75bd
2 Answers
Let's rewrite this to make our life easier:
Now take the first derivative:
For the second derivative, derive the first derivative.
The second derivative is negative, meaning that the graph is a maximum.
Explanation:
"assuming "y=sqrt(2-3x^2)assuming y=√2−3x2
"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)