find y if y = ln (x^4 + 1)^3/2. y =

Explanation:

Step1: Use power - rule of logarithms

By the power - rule of logarithms $\ln a^b = b\ln a$, we can rewrite $y=\ln(x^{4}+1)^{\frac{3}{2}}$ as $y = \frac{3}{2}\ln(x^{4}+1)$.

Step2: Apply the chain - rule for differentiation

The derivative of $\ln u$ with respect to $x$ is $\frac{u'}{u}$ by the chain - rule. Let $u=x^{4}+1$, then $u' = 4x^{3}$. The derivative of $y=\frac{3}{2}\ln(x^{4}+1)$ is $y'=\frac{3}{2}\times\frac{4x^{3}}{x^{4}+1}$.

Step3: Simplify the expression

$y'=\frac{3\times4x^{3}}{2(x^{4}+1)}=\frac{6x^{3}}{x^{4}+1}$.

Answer:

$\frac{6x^{3}}{x^{4}+1}$