Question

find dy.
y = 4 ln (2 + x^{4})
dy = □ 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:

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