Answer to Question #305335 in Calculus for justt

Question

Using the Chain Rule, find the 𝑑𝑦/𝑑𝑥 and express the final answer in term of x.

𝑦 = 2𝑢/𝑢²−1 , u=x²

Answer 1

Solution

Given that

$y=\frac{2u}{u^{2}-1} ... (1)\\ \\ u=x^{2} ... (2)$

From (1),

${\color{Blue} y=\frac{2u}{u^{2}-1}}\\ \frac{dy}{du}=\frac{(u^{2}-1)\frac{d}{du}(2u)-(2u)\frac{d}{du}(u^{2}-1)}{(u^{2}-1)^{2}}\\ \frac{dy}{du}=\frac{(u^{2}-1)(2)-(2u)(2u)}{(u^{2}-1)^{2}}\\ \frac{dy}{du}=\frac{(2u^{2}-2)-(4u^{2})}{(u^{2}-1)^{2}}\\ \frac{dy}{du}=\frac{(-2u^{2}-2)}{(u^{2}-1)^{2}}\\ \frac{dy}{du}=\frac{-2(u^{2}+1)}{(u^{2}-1)^{2}} ... (3)\\$

Now from (2)

${\color{Blue}u=x^{2}}\\ \frac{du}{dx}=\frac{d}{dx}(x^{2})\\ \frac{du}{dx}=2x ... (4)$

According to the chain rule

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

Using (3) and (4)

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\\ \\ \\ \frac{dy}{dx}=\frac{-2(u^{2}+1)}{(u^{2}-1)^{2}}\times (2x)\\ \\ \\ \frac{dy}{dx}=\frac{-4x(u^{2}+1)}{(u^{2}-1)^{2}}\\$

Hence according to the chain rule,

${\color{Blue}\frac{dy}{dx}=\frac{-4x(u^{2}+1)}{(u^{2}-1)^{2}}}\\$