1. ## 3 Interesting Questions

Stuck on these questions would appreciate the help very much!!!

2. Originally Posted by student3141

Stuck on these questions would appreciate the help very much!!!

This looks like homerwork and you show no self work at all...what've you done so far?

Tonio

3. ive been trying but im tottaly stuck dont know where to begin! or how to approach it

4. Originally Posted by student3141
ive been trying but im tottaly stuck dont know where to begin! or how to approach it
Have you reviewed your class notes and textbook for similar examples that you can follow?

5. Ok i did question 7

8) i tried to get the mvt for both sides, do i treat the ln(x) and the RHS as seperate? and do i then juss find the mvt for both sides? cause when i try and get the mvt for the RHS i get 3/8 i think thats wrong isnt it supposed to be between 1 and 2??

9) This question i kno its implicit differentiation but what should i do with the A B C ???
I cant seem to find similar questions in text book... help needed!!

6. Treat $\displaystyle A,B,C$ as constants.

7. For question 9 this is wat i did help!

2Ax + 2By(dy/dx) = 0

Dy/Dx = Ax/By

D^2y/Dx^2 = ByA - BAx(dy/dx) / (B^2).(y^2)

= BA(y-x(dy/dx)) / (B^2).(y^2)

whats going on????? did i do something wrong?

8. $\displaystyle 2A\,x + 2By\,\frac{dy}{dx} = 0$

$\displaystyle 2B\,y\,\frac{dy}{dx} = -2A\,x$

$\displaystyle \frac{dy}{dx} = -\frac{2A\,x}{2B\,y}$

$\displaystyle \frac{dy}{dx} = -\frac{A\,x}{B\,y}$.

To find $\displaystyle \frac{d^2y}{dx^2}$ you need to use the Quotient Rule... Also remember that you already know what $\displaystyle \frac{dy}{dx}$ is so you can substitute it in...

9. btw thanks for the help

Thats quotient rule so:

dy/dx = - Ax/By

= - (ByA - AxB(dy/dx))/(B^2).(y^2)

= BA ( y - x(dy/dx))/(B^2).(y^2)

i get stuck over here

10. Be more careful with what you write, in this step you're finding $\displaystyle \frac{d^2y}{dx^2}$, not $\displaystyle \frac{dy}{dx}$.

Also, you forgot to bring the negative down into your last step.

Now, assuming you had brought the negative down...

$\displaystyle -\frac{AB\left(y - x\,\frac{dy}{dx}\right)}{B^2y^2} = -\frac{A\left(y - x\,\frac{dy}{dx}\right)}{B\,y^2}$

$\displaystyle = -\frac{A\left[y - x\left(-\frac{A\,x}{B\,y}\right)\right]}{B\,y^2}$

11. AHHH yess i get it now thank you so much!
Btw did you use a program to write that working out ? looks much more neater than what i was doing!

For question 8

(3/2)x - (1/4)x^2 - (5/4)

F(2)-F(1) / 2-1 = 3/4

dy/dx = 3/2 - (1/2)x

3/4 = 3/2 - (1/2)c

c= 3/2

ok so do i now juss work out the mvt of ln(x)???

12. I'm using the LaTeX compiler that is built into the forum.

See the LaTeX help subforum to learn how to use it.