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
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
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