Question

I really need help!

solve the equation for y' = 0.

y = 5x^2 + x -1


I dont know how to do this...

I know it has to deal with the CHAIN RULE

Answer 1

it should be -1/10.
if you take the derivative of the function Y then y prime will be 10x+1 and if you set y prime equal to zero then x should be -1/10

Answer 2

wait so after you get the derivative of y which is 10x+1 you do substitute zero???

Answer 3

actually it doesn't require the chain rule.

all you need to do is

1. find the derivative

2. set the derivative equal to zero

3. solve the derivative for x

Answer 4

your derivative is correct

y' = 10x + 1

now set it to zero

0 = 10x + 1

and solve for x

Answer 5

y=5x^2+x-1
y'=dy/dx=10x+1=0
10x=-1,x=-1/10,
y=5(-1/10)^2+(-1/10)-1=5/100-1/10 -1=(5-10-100)/100=-105/100=-21/20
x=-1/10, y=-21/20

Answer 6

=∫▒〖logx xdx〗 ∫▒〖xlogx dx〗
=logxx^2/2-∫▒〖1/x x^2/2〗 dx
=logx x^2/2-1/2 ∫▒xdx
=logx x^2/2 -1/2 x^2/2 +c
=logx *x^2/2 -x^2/4+c
=∫▒〖logx xdx〗 ∫▒〖xlogx dx〗
=logxx^2/2-∫▒〖1/x x^2/2〗 dx
=logx x^2/2-1/2 ∫▒xdx
=logx x^2/2 -1/2 x^2/2 +c
=logx *x^2/2 -x^2/4+c
∫▒〖logx xdx〗=∫▒〖xlogx dx〗
=logxx^2/2-∫▒〖1/x x^2/2〗 dx
=logx x^2/2-1/2 ∫▒xdx
=logx x^2/2 -1/2 x^2/2 +c
=logx *x^2/2 -x^2/4+c