differentiate. g(x)=ln(x^4 + 1)^3 g(x)=

Explanation:

Step1: Use power - rule of logarithms

First, use the power - rule of logarithms $\ln(a^b)=b\ln(a)$. So, $g(x)=\ln((x^{4}+1)^{3}) = 3\ln(x^{4}+1)$.

Step2: Apply the chain - rule

The derivative of $y = 3\ln(u)$ with respect to $x$ (where $u=x^{4}+1$) is given by the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. The derivative of $y = 3\ln(u)$ with respect to $u$ is $\frac{3}{u}$, and the derivative of $u=x^{4}+1$ with respect to $x$ is $4x^{3}$.

Step3: Substitute $u$ back

Substitute $u = x^{4}+1$ into $\frac{3}{u}\cdot4x^{3}$. We get $g^{\prime}(x)=\frac{3\cdot4x^{3}}{x^{4}+1}=\frac{12x^{3}}{x^{4}+1}$.

Answer:

$\frac{12x^{3}}{x^{4}+1}$