Help Needed With Differentiation!

[rozzer123]

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:



Here is my working thus far:

Let [MATH]u = 1+bx[/MATH], then [MATH]u' = b[/MATH].

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

Therefore, [MATH]\dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{-a}{2}u^{-3/2} * b[/MATH].

Substitute u in [MATH]\dfrac{dy}{dx}[/MATH];

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

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

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

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

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

(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

 

[Steven G]

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]

Obtain the secong equation by substituting x=1 and y=1 into the original eqtn.

 

[Steven G]

Obtain the secong equation by substituting x=1 and y=1 into the original eqtn.

You're so nice.

 

[rozzer123]

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]

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!

 

[Steven G]

So what results did you get for a and b. Please post back. It will help other students with similar problems.

 

[rozzer123]

So what results did you get for a and b. Please post back. It will help other students with similar problems.

Sincere apologies for such a late reply.

As Harry_the_cat kindly advised, I substituted the x=1 and y=1 values back into the original equation. This led me to my next thinking:

[MATH]1=\frac{a}{\sqrt{1+b}}[/MATH][MATH]a=\sqrt{1+b}[/MATH]By substituting the formula we have just calculated into the other equation,
[MATH]-\frac{3}{8}=-\frac{b\sqrt{1+b}}{2\sqrt{(1+b)^3}}[/MATH][MATH]6\sqrt{(1+b)^3}=8b\sqrt{1+b}[/MATH][MATH]36(1+b)^3=64b^2(1+b)[/MATH][MATH]-28b^3+44b^2+108b+36=0[/MATH][MATH]b=-1[/MATH] or [MATH]b=-\frac{3}{7}[/MATH] or [MATH]b=3[/MATH]
The first two solutions do not stand true when substituted into [MATH]-\frac{3}{8}=-\frac{b\sqrt{1+b}}{2\sqrt{(1+b)^3}}[/MATH].

The last solution stands true when substituted. Therefore, the first two solutions are rejected and the last is accepted.

When [MATH]b=3[/MATH],

[MATH]a=\sqrt{1+3}=\sqrt{4}=2[/MATH].

Thus, [MATH]a=2[/MATH] and [MATH]b=3[/MATH].

I hope I'm correct!

 

[Steven G]

Please reduce before you square both sides, please.

 

[JeffM]

[MATH]y = \dfrac{2}{\sqrt{1 + 3x}} \implies \dfrac{2}{\sqrt{1 + 3(1)}} = \dfrac{2}{\sqrt{4}} = 1 \ \checkmark.[/MATH]
[MATH]y = \dfrac{2}{\sqrt{1 + 3x}} = 2(1 + 3x)^{-1/2} \implies y^2 = 4(1 + 3x)^{-1} \implies \\ 2yy' = -(1)(4)(3)(1 + 3x)^{-2} \implies \\ y' * 2 * \dfrac{2}{(1 + 3x)^{1/2}} = - \dfrac{12}{(1 + 3x)^2} \implies y' = - \dfrac{3\sqrt{1 + 3x}}{(1 + 3x)^2}.\\ \therefore - \dfrac{3\sqrt{1 + 3(1)}}{(1 + 3(1))^2} = - \dfrac{3 * \sqrt{4}}{4^2} = - \dfrac{6}{16} = - \dfrac{3}{8}. \ \checkmark[/MATH]Looks good to me.

By the way, it might have been slightly easier to get rid of a before doing any differentiation.

[MATH]y = 1 \text { if } x = 1 \text { and } y = \dfrac{a}{\sqrt{1 + bx}} \implies \\ 1 = \dfrac{a}{\sqrt{1 + b(1)}} \implies a = \sqrt{1 + b} \implies y = \dfrac{\sqrt{1 + b}}{\sqrt{1 + bx}}.[/MATH]