1. ## Need Derivitive Help

Well, i have two problems, and i'm not really sure if the first is derivitive or not. Please help!

1. At time t a particle is moving along the x-axis is at position x. The relationship between t and x is given by tx = x(squared) + 8. At x = 2, the velocity of the particle is:

A. 1 B. 2 C. 6 D. -2 E. -1

I don't know what to do here. I tried just plugging 2 into the equation, solving to t, and doing velocity equals distance over time, but that didn't work.

2. If x(squared) + 2xy - 3y = 3, dy/dx of x = 2 equals:

A. 1 B. 2 C. -2 D. 10/3 E. -1

I tried the product rule of 2x and y, but it didn't work out right.

Actually there is another problem that's even harder

If d/dx of f(x) = g(x) and d/dx of g(x) = f(3x), then d(squared)/dx(squared) of f(x(squared)) equals:

A. 4x(squared)f(3x(squared)) + 2g(x(squared)) B. f(3x(squared)) C. f(x^4) D. 2xf(3x(squared)) + 2g(x(squared)) E. 2xf(3x(squared))

I'm pretty sure its not B or C, but i really have no idea how to do this problem. My teacher did mention using the chain rule as a hint, but i stil really don't get it.

2. Hello, Blue Griffin!

1. At time t a particle is moving along the x-axis is at position x.
The relationship between t and x is given by: . $tx \:= \:x^2 + 8$

At x = 2, the velocity of the particle is:

. . $(A)\;1\qquad(B)\;2\qquad(C)\;6\qquad(D)\;-2\qquad(E)\;-1$
We have: . $t \;=\;x + 8x^{-1}$

. . Then: . $\frac{dt}{dx} \:=\:1-8x^{-2} \:=\:\frac{x^2-8}{x^2}$

. . Hence: . $\frac{dx}{dt} \:=\:\frac{x^2}{x^2-8}$

At $x = 2\!:\;\frac{dx}{dt} \;=\;\frac{2^2}{2^2-8} \;=\;\frac{4}{\text{-}4} \;=\;-1$ . Answer (E)

$2)\text{ If }\:x^2 + 2xy - 3y \:=\: 3,\;\text{ find: }\frac{dy}{dx}\;\text{ at }x = 2.$
. . $(A)\;1\qquad(B)\;2\qquad(C)\;-2\qquad(D)\;\frac{10}{3}\qquad(E)\;-1$
Differentiate implicitly: . $2x + 2x\!\cdot\!\frac{dy}{dx} + 2y - 3\!\cdot\!\frac{dy}{dx}\:=\:0$

. . $2x\!\cdot\!\frac{dy}{dx} - 3\!\cdot\!\frac{dy}{dx} \:=\:-2x-2y$

. . $(2x-3)\!\cdot\!\frac{dy}{dx}\:=\:-2(x+y)$

. . $\frac{dy}{dx}\:=\:\frac{-2(x+y)}{2x-3} \:=\:\frac{2(x+y)}{3-2x}$

When $x = 2$, the original equation is: . $4 + 4y - 3y \:=\:3\quad\Rightarrow\quad y = \text{-}1$

Therefore: . $\frac{dy}{dx} \:=\:\frac{2+(\text{-}1)}{3-2(2)} \:=\:-1$ . Answer (E)

$3)\;\text{If }\:f'(x)\:=\:g(x)\;\text{ and }\;g'(x)\:=\:f(3x),\;\text{find: }f''(x^2)$

$(A)\;4x^2\!\cdot\!f(3x^2) + 2\!\cdot\!g(x^2)\quad B)\;f\left(3x^2\right)\quad (C)\;f\left(x^4\right)\quad (D)\;2x\!\cdot\!f(3x^2)\: +$ $2\!\cdot\!g(x^2)\quad (E)\;2x\!\cdot\!f(3x^2)$
We have: . $y\;=\;f(x^2)$

Then: . $y' \;=\;f'(x^2)\cdot2x$ . . . . but $f'(x^2) \,=\,g(x^2)$

So we have: . $y' \:=\:2x\cdot g(x^2)$

Product Rule: . $y'' \:=\:2x\!\cdot\!g'(x^2)\!\cdot\!2x + 2\!\cdot\!g(x^2)$ . . . . but $g'(x^2) \:=\:f(3x^2)$

Therefore: . $y'' \;=\;4x^2\!\cdot\!f(x^2) + 2\!\cdot\!g(x^2)$ . Answer (A)

3. Thanks a lot. It all makes sense now, lol.