Question

need help on the attachment

Answer 1

I have no idea what - Oh.

Answer 2

you don't know how to work the problem

Answer 3

No, I didn't know what the attachment was. Now I see. It's a related rate problem. Sec.

Answer 4

y = x^2 - 2x

Let's take the derivative of both sides.
d/dt [y] = d/dx [x^2 - 2x]
dy/dt = 2x dx/dt - 2 dx/dt

Plug in known values.

dy/dt = 2 (3) (3) - 2 (3)
dy/dt = 12

I hope this is correct - a little rusty with this type of problem. Reply back!

Answer 5

yes its right thanks!

Answer 6

Whoops. Typo in line 4 (not counting spaces). It should read:
d/dt [y] = d/dt [x^2 - 2x]

I put d/dx by mistake. You're welcome!

Answer 7

oh, ok could you help me with one more problem

Answer 8

Also why did you take derivative when all you need to do is plug in the value of 3 that they give you?

Answer 9

Sure, I can try to help with another.

Answer 10

I had to find the derivative of y=x^2 - 2x with respect to t in order to do anything at all. It asked for dy/dt, not y.

Answer 11

but why did you times the 2X by 3 two times

Answer 12

Okay, you're asking about what I did after this, correct?

dy/dt = 2(x) (dx/dt) - 2 (dx/dt)

The question stated that x = 3 and dx/dt = 3, so I plugged in 3 for both x and dx/dt.

dy/dt = 2 (3) (3) - 2 (3)

Answer 13

oh, ok

Answer 14

\[f(x) = (x^2 - 2) / \sqrt{x ^{2}-1}\]

I prefer to do the chain and product rule as opposed to the quotient rule, by doing this.

f(x) = (x^2 - 2) (x^2 - 1)^-1/2

Now, use product rule. Chain rule also comes into play.

f'(x) = 2x(x^2 - 1)^(-1/2) + (x^2 - 2) (2x) (-1/2) (x^-1)^(-3/2)

Simplify from there. Or use this. http://www4d.wolframalpha.com/Calculate/MSP/MSP25319hc42hg3hdfa4b400000f7hgha31eg7ie17?MSPStoreType=image/gif&s=17&w=465&h=1055

Answer 15

so what the final answer?

Answer 16

y=(3)^(2)-2(3)

Expand the exponent (2) to the expression.
y=(3^(2))-2(3)

Squaring a number is the same as multiplying the number by itself (3*3). In this case, 3 squared is 9.
y=(9)-2(3)

Remove the parentheses from the numerator.
y=9-2(3)

Multiply -2 by 3 in the numerator.
y=9+(-2*3)

Multiply -2 by 3 to get -6.
y=9+(-6)

Move the minus sign from the numerator to the front of the expression.
y=9-6

Subtract 6 from 9 to get 3.
y=3

Answer 17

or solve for x,
y=x^(2)-2x

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
x^(2)-2x=y

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
x^(2)-2x-y=0

Use the quadratic formula to find the solutions. In this case, the values are a=1, b=-2, and c=-1y.
x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

Use the standard form of the equation to find a, b, and c for this quadratic.
a=1, b=-2, and c=-1y

Substitute in the values of a=1, b=-2, and c=-1y.
x=(-(-2)\~((-2)^(2)-4(1)(-1y)))/(2(1))

Multiply -1 by -2 in the numerator.
x=((-1*-2)\~((-2)^(2)-4(1)(-1y)))/(2(1))

Multiply -1 by -2 to get 2.
x=((2)\~((-2)^(2)-4(1)(-1y)))/(2(1))

Remove the parentheses from the numerator.
x=(2\~((-2)^(2)-4(1)(-1y)))/(2(1))

Simplify the section inside the radical.
x=(2\2~(y+1))/(2(1))

Simplify the denominator of the quadratic formula.
x=(2\2~(y+1))/(2)

First, solve the + portion of \.
x=(2+2~(y+1))/(2)

Simplify the expression to solve for the + portion of the \.
x=1+~(y+1)

Next, solve the - portion of \.
x=(2-2~(y+1))/(2)

Simplify the expression to solve for the - portion of the \.
x=1-~(y+1)

The final answer is the combination of both solutions.
x=1+~(y+1),1-~(y+1)