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 \(\displaystyle u = 1+bx\), 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!)

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 \(\displaystyle u = 1+bx\), 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

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