# Differentiation structured questions

• Dec 14th 2010, 01:36 AM
arccos
Differentiation structured questions
Hey,

I need help with some questions that I have been stuck with on my crash course to teach myself calculus before school starts.

1) If $y = x +\sqrt{x^2 -9}$ , show that

$\displaystyle (x^2 -9) \dfrac{d^2y}{dx^2} + x \dfrac{dy}{dx} - y = 0$

I have tried to solve for the derivatives of y and I derived extremely complicated fractions with powers of 3/2 and such and I cannot simplify the equation to = 0.

2)
a) $\dfrac {d}{dx} \dfrac{3x-4}{x+7} = \dfrac {25}{(x+7)^2}$ (found)

b) Using your answer in (a) , differentiate the following with respect to x
$\displaystyle y = sin(\dfrac{3x-4}{x+7})^8$

For this, I am unsure of what to let be u so that I can use the chain rule in the form
$\dfrac{du}{du} \, \dfrac{du}{dx}$ or $\dfrac{dsinu}{du} \, \dfrac{du}{dx}$
I have dealt with trigo functions like sin^8(x) and sin(8x) but never in the form that the question gave (8th power only applies to the fraction and not the whole function) and I do not know how to proceed.

Any help at all is appreciated :)
• Dec 14th 2010, 02:30 AM
Ackbeet
Well, what did you get for the derivatives in 1)?

For 2) b), you have to use the chain rule from the outside in, and you're going to have to use it twice (having done part a means you don't have to do it three times, like you normally would). Here's the first step:

$\displaystyle \frac{dy}{dx} = \cos\left(\left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right)\;\frac{d}{dx}\left(\ left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right).$

Thank you for the sentence clarifying what the exponent 8 applies to, but if you include extra parentheses like I've done here, you don't need to bother.

Can you keep going from here?
• Dec 14th 2010, 04:46 AM
arccos
Quote:

Originally Posted by Ackbeet
Well, what did you get for the derivatives in 1)?

For 2) b), you have to use the chain rule from the outside in, and you're going to have to use it twice (having done part a means you don't have to do it three times, like you normally would). Here's the first step:

$\displaystyle \frac{dy}{dx} = \cos\left(\left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right)\;\frac{d}{dx}\left(\ left(\dfrac{3x-4}{x+7}\right)^{\!\!8}\right).$

Thank you for the sentence clarifying what the exponent 8 applies to, but if you include extra parentheses like I've done here, you don't need to bother.

Can you keep going from here?

Thanks for the tip. 2 is solved.
For qn 1 ,
$\dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{x^2 -9}}$

$\dfrac{d^2y}{dx^2} = \dfrac{1}{\sqrt{x^2-9}} - \dfrac{x^2}{(x^2-9)^{\frac{3}{2}}}$

All the weird exponents make simplifying hard. Have i gone wrong with the deriviation?
• Dec 14th 2010, 05:32 AM
HallsofIvy
$A^{3/2}= A(A^{1/2})= A\sqrt(A)$ so that could be written
$\frac{d^2y}{dx^2}= \frac{1}{\sqrt{x^2- 9}}- \frac{x^2}{(x^2-9)\sqrt{x^2- 9}}$
and then get "common denominators by multiplying the numerator and denominator of the first fraction by $x^2- 9$:
$\frac{d^2y}{dx^2}= \frac{x^2- 9}{(x^2-9)\sqrt{x^2- 9}}- \frac{x^2}{(x^2- 9)\sqrt{x^2- 9}}$
$\frac{d^2y}{dx^2}= -\frac{9}{(x^2- 9)\sqrt{x^2- 9}}$
• Dec 14th 2010, 05:56 AM
arccos
$\dfrac{-9}{\sqrt{x^2-9}} + \dfrac{x^2}{\sqrt{x^2-9}} ....$

In the end, the 9 and x^2 canceled out,I crossed out sqrt x^2-9 from both the denominator and numerator but im left with -x...

Sorry about the shortcut,i screwed up my latex. (Speechless)

EDIT:
Oh I have solved the problem. The thing I overlooked was to not simplify the numerator but to group them together.
$\dfrac{-9}{\sqrt{x^2-9}} + \dfrac{x^2}{\sqrt{x^2-9}}$'s numerator can be regrouped to x^2 - 9 which cancels out with the other x^2 - 9 , leaving 0 which divided by anything,is still = 0 = RHS.

Thanks for the help anyway guys. I appreciate it :)
• Dec 15th 2010, 06:11 AM
TheCoffeeMachine
Differentiating and rearranging gives:

$\displaystyle {y = x +\sqrt{x^2 -9}}} \Rightarrow y' = 1+\frac{x}{\sqrt{x^2-9}} \Rightarrow y' -\frac{x}{\sqrt{x^2-9}}= 1$

But $x = y-\sqrt{x^2-9}$, so we have:

$\displaystyle \Rightarrow y'-\frac{y-\sqrt{x^2-9}}{\sqrt{x^2-9}} = 1 \Rightarrow y'-\frac{y}{\sqrt{x^2-9}} = 0 ~ ~ \;\; (\star)$

Differentiate this, noting of course that:

$\displaystyle \left(\frac{y}{\sqrt{x^2-9}}\right)' = \frac{y'\sqrt{x^2-9}-y\left(\sqrt{x^2-9}\right)'}{(x^2-9)} = \frac{y'\left(x^2-9\right)-yx}{(x^2-9)\sqrt{x^2-9}}$

So we have:

$\displaystyle y''-\frac{y'\left(x^2-9\right)-yx}{(x^2-9)\sqrt{x^2-9}} = 0 \Rightarrow y''-\frac{y'(x^2-9)}{(x^2-9)\sqrt{x^2-9}}-x\frac{y}{(x^2-9)\sqrt{x^2-9}} = 0
$

$\displaystyle \Rightarrow (x^2-9)y''-\frac{y'(x^2-9)}{\sqrt{x^2-9}}-x\frac{y}{\sqrt{x^2-9}} = 0 \Rightarrow (x^2-9)y''-y'\sqrt{x^2-9}-x\frac{y}{\sqrt{x^2-9}} = 0$

But from $\left(\star\right)$ we have $y'\sqrt{x^2-9} = y$ and $\frac{y}{\sqrt{x^2-9}} = -y'$, so indeed $\displaystyle \boxed{(x^2-9)y''+xy'-y = 0}.$