find dy.\ny = 4 ln (2 + x^{4})\ndy = □ dx

Explanation:

Step1: Differentiate ( y = 4\ln(2 + x^{4}) ) using the chain rule

The chain rule states that if ( y = f(g(x)) ), then ( y^\prime=f^\prime(g(x))\cdot g^\prime(x) ). For ( y = 4\ln(u) ) where ( u = 2 + x^{4} ), the derivative of ( \ln(u) ) with respect to ( u ) is ( \frac{1}{u} ), and the derivative of ( u = 2 + x^{4} ) with respect to ( x ) is ( 4x^{3} ). First, ( \frac{dy}{du}=4\times\frac{1}{u}) (since the derivative of ( y = 4\ln(u) ) with ( u = 2 + x^{4} )), and ( \frac{du}{dx}=4x^{3} ). By the chain rule ( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). Substitute ( u = 2 + x^{4} ) into ( \frac{dy}{du} ), we get ( \frac{dy}{du}=\frac{4}{2 + x^{4}} ). Then ( \frac{dy}{dx}=\frac{4}{2 + x^{4}}\cdot4x^{3}).

Step2: Simplify the expression

Multiply the constants and variables: ( \frac{dy}{dx}=\frac{16x^{3}}{2 + x^{4}} ). Since ( dy=\frac{dy}{dx}dx ), then ( dy=\frac{16x^{3}}{2 + x^{4}}dx ).

Answer:

( dy=\frac{16x^{3}}{2 + x^{4}}dx )