1. ## Review Question

Hi
I have some question i am having trouble with:
1)If the angle between the lines which passes through the points (-2,3) and (4,0) is:
A $2y=x+4$
B $y=\frac{-x}{2}-2$
C $2y+x=4$
D $y=\frac{x}{2}-2$
E $2y-x=4$

2)Find the derivative of $(4x+\frac{9}{x})^2$ and values of x at which the derivative is zero

3)Let f be differentiable for all values of x in [2,0]. The graph with equation y=f(x) has a maximum point at (1,3). The equation of the tangent at (1,3) is:
A $x+3y=0$
B $x=1$
C $y=3$
D $x-3y=0$
E $3x+y=0$

P.S

2. Originally Posted by Paymemoney
Hi
I have some question i am having trouble with:
1)If the angle between the lines which passes through the points (-2,3) and (4,0) is:

Have a look at this post...

http://www.mathhelpforum.com/math-he...-problems.html

3. Originally Posted by pickslides
thanks, i know how to do this now

4. Originally Posted by Paymemoney

2)Find the derivative of $\left(4x+\frac{9}{x}\right)^2$ and values of x at which the derivative is zero
Lets say $y = \left(4x+\frac{9}{x}\right)^2$

Using the chain rule

$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

Make $u = 4x+\frac{9}{x}$ and $y = u^2$

Now find $\frac{dy}{du}$ and $\frac{du}{dx}$

5. Originally Posted by pickslides
Lets say $y = \left(4x+\frac{9}{x}\right)^2$

Using the chain rule

$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

Make $u = 4x+\frac{9}{x}$ and $y = u^2$

Now find $\frac{dy}{du}$ and $\frac{du}{dx}$
This is how far i got until i got stuck.
$2(4x+\frac{9}{x})(4+\frac{9}{x^2})$

$8(4x+\frac{9}{x})+\frac{9}{x^2}$

$32x+\frac{72}{x}-\frac{9}{x^2}$

$\frac{32x^3}{x^2}+\frac{72x}{x^2}-\frac{9}{x^2}$

6. $u = 4x+\frac{9}{x} = 4x+9x^{-1}$ and $y = u^2$

$\frac{dy}{du} = 2u = 2\left(4x+\frac{9}{x}\right)$

$\frac{du}{dx} = 4-9x^{-2}$

$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

$\frac{dy}{dx} = 2\left(4x+\frac{9}{x}\right)\times \left(4-9x^{-2}\right)$

Now expnad using the F.O.I.L method. Do you know this method?

7. Originally Posted by pickslides
$u = 4x+\frac{9}{x} = 4x+9x^{-1}$ and $y = u^2$

$\frac{dy}{du} = 2u = 2\left(4x+\frac{9}{x}\right)$

$\frac{du}{dx} = 4-9x^{-2}$

$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

$\frac{dy}{dx} = 2\left(4x+\frac{9}{x}\right)\times \left(4-9x^{-2}\right)$

Now expnad using the F.O.I.L method. Do you know this method?
oh ok yeh i know what the method is now.