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!!
$\displaystyle \displaystyle 2A\,x + 2By\,\frac{dy}{dx} = 0$
$\displaystyle \displaystyle 2B\,y\,\frac{dy}{dx} = -2A\,x$
$\displaystyle \displaystyle \frac{dy}{dx} = -\frac{2A\,x}{2B\,y}$
$\displaystyle \displaystyle \frac{dy}{dx} = -\frac{A\,x}{B\,y}$.
To find $\displaystyle \displaystyle \frac{d^2y}{dx^2}$ you need to use the Quotient Rule... Also remember that you already know what $\displaystyle \displaystyle \frac{dy}{dx}$ is so you can substitute it in...
Be more careful with what you write, in this step you're finding $\displaystyle \displaystyle \frac{d^2y}{dx^2}$, not $\displaystyle \displaystyle \frac{dy}{dx}$.
Also, you forgot to bring the negative down into your last step.
Now, assuming you had brought the negative down...
$\displaystyle \displaystyle -\frac{AB\left(y - x\,\frac{dy}{dx}\right)}{B^2y^2} = -\frac{A\left(y - x\,\frac{dy}{dx}\right)}{B\,y^2}$
$\displaystyle \displaystyle = -\frac{A\left[y - x\left(-\frac{A\,x}{B\,y}\right)\right]}{B\,y^2}$
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)???