Help Needed With Differentiation!

rozzer123

New member
Joined
Sep 25, 2019
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7
Hello,

I've been going through my textbook and I am currently working through the Differentiation topic. I feel like I'm getting a good grip with the concepts (such as the chain rule, etc.), but I've been stuck on this problem for quite some time.

Here is the question:

Screen Shot 2020-03-30 at 8.04.59 PM.png

Here is my working thus far:

Let \(\displaystyle u = 1+bx\), then \(\displaystyle u' = b\).

Hence, \(\displaystyle y = au^{-1/2}\) and \(\displaystyle y' = \)\(\displaystyle \dfrac{-a}{2}\)\(\displaystyle u^{-3/2}\).

Therefore, \(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{-a}{2}u^{-3/2} * b\).

Substitute u in \(\displaystyle \dfrac{dy}{dx}\);

\(\displaystyle \dfrac{dy}{dx} = \)\(\displaystyle \dfrac{-a}{2}\)\(\displaystyle (1+bx)^{-3/2} * b\).

Since \(\displaystyle \dfrac{dy}{dx}\) = \(\displaystyle \dfrac{-3}{8}\) at point (1,1)

\(\displaystyle \dfrac{-3}{8}\) = \(\displaystyle \dfrac{-a}{2}(1+bx)^{-3/2} * b\).

Therefore, substituting x as 1: \(\displaystyle \dfrac{-3}{8}\) = \(\displaystyle \dfrac{-a}{2}(1+b)^{-3/2} * b\).

So, \(\displaystyle \dfrac{-3}{8}\) = \(\displaystyle \dfrac{-ab}{2(1+b)^{-3/2}}\).

(I know the index rules regarding the {-3/2}, I just couldn't find a way to type in the root symbol in the coding program :3... even from there, I have no idea where to go!)

Am I approaching the problem correctly? I feel like, after this step, I've been spiraling around - never able to arrive at the correct solution. I've attempted solving for the polynomial roots, using simultaneous equations.. but all have stopped at a dead end.

Could someone help me out? Any advice will genuinely be appreciated :)
 
Last edited:

Jomo

Elite Member
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Dec 30, 2014
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7,035
Your last equation is not correct.

After that you should get 2 equations both involving a and b. Solve that system to find a and b
 

Harry_the_cat

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Mar 16, 2016
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Obtain the secong equation by substituting x=1 and y=1 into the original eqtn.
 

Jomo

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rozzer123

New member
Joined
Sep 25, 2019
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Your last equation is not correct.

After that you should get 2 equations both involving a and b. Solve that system to find a and b
Alright. Thank you so much!
 

rozzer123

New member
Joined
Sep 25, 2019
Messages
7
S
Obtain the secong equation by substituting x=1 and y=1 into the original eqtn.
Thank you. I didn't think about looking back at the original equation to obtain the second equation!
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
7,035
So what results did you get for a and b. Please post back. It will help other students with similar problems.
 
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