Question #305335

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

𝑦 = 2𝑢/𝑢²−1 , u=x²


1
Expert's answer
2022-03-03T17:49:51-0500

Solution


Given that


y=2uu21...(1)u=x2...(2)y=\frac{2u}{u^{2}-1} ... (1)\\ \\ u=x^{2} ... (2)


From (1),


y=2uu21dydu=(u21)ddu(2u)(2u)ddu(u21)(u21)2dydu=(u21)(2)(2u)(2u)(u21)2dydu=(2u22)(4u2)(u21)2dydu=(2u22)(u21)2dydu=2(u2+1)(u21)2...(3){\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)

u=x2dudx=ddx(x2)dudx=2x...(4){\color{Blue}u=x^{2}}\\ \frac{du}{dx}=\frac{d}{dx}(x^{2})\\ \frac{du}{dx}=2x ... (4)


According to the chain rule

dydx=dydu×dudx\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}



Using (3) and (4)


dydx=dydu×dudxdydx=2(u2+1)(u21)2×(2x)dydx=4x(u2+1)(u21)2\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,

dydx=4x(u2+1)(u21)2{\color{Blue}\frac{dy}{dx}=\frac{-4x(u^{2}+1)}{(u^{2}-1)^{2}}}\\



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