Question

Find dy/dx given that y² =x(3-x)². State the values of x for which dy/dx is: (a) zero (b) infinite.

answer to (a) ± (3/2√x-(3/2)√x)
answer to (b) is 1

HI CAN YOU SHOW ME STEP BY STEP HOW TO GET X=± (3/2√x-(3/2)√x). AS AM GETTING A DIFFERENT ANSWER (x^2=9/12x)

PS
and part b how do i get the 1 thanks

Answer 1

is that a z in the exponent or a 2?

Answer 2

its 2

Answer 3

Ok.

Answer 4

Show your attempt at finding dy/dx first.

Answer 5

ok using implicit differentiation

Answer 6

and also function of a function rule for (3-x)²

Answer 7

i get 2ydy/dx=-2x (3-x)²+ (3-x)².1

Answer 8

sorry it should be 2ydy/dx= 2x (3-x)+(3-x)²

Answer 9

when you expand and set dy/dx=0 you get 0=-12x-x²+9

Answer 10

getting x²=9/12x

Answer 11

which doesn't seem correct

Answer 12

y² =x(3-x)²
Using the chain rule and the product rule together:
\[ 2yy' = 1*x(3-x)^2 + 2x(3-x) * (-1) \]
\[\ y' = \frac {x(3-x)^2 - 2x(3-x)} {2} \]

Answer 13

ok

Answer 14

x(3-x)^2
(x)(2(3-x)*-1)+(1(3-x)^2)
(x)(6-2x-1)+(3-x)^2
6x-2x^2-x +(3-x)^2
-2x^2+5x+(3-x)^2

Answer 15

ok

Answer 16

expanding (3-x)^2 we get 9-6x+x^2

Answer 17

so we get -x^2-x+9

Answer 18

is that correct?

Answer 19

Using the product rule and chain rule, yes. I am trying to figure out why shamil has x's in front of the parenthesis

Answer 20

I think shamil has written x by mistaken

Answer 21

also, in 2nd step,
denominator should be 2y

Answer 22

\[y'=\frac{ 3(x^2-4x+3) }{ 2y }\]

Answer 23

ok

Answer 24

so do we substitute y into the denominator?