use logarithmic differentiation to find the derivative of y with respect to the given independent variable…

Explanation:

Step1: Apply product - rule of logarithms

By the product - rule of logarithms $\ln(abc)=\ln a+\ln b+\ln c$, if $y = t(t + 7)(t + 5)$, then $\ln y=\ln t+\ln(t + 7)+\ln(t + 5)$.

Step2: Differentiate both sides with respect to $t$

The derivative of $\ln y$ with respect to $t$ is $\frac{1}{y}\cdot y'$ (by the chain - rule). The derivative of $\ln t$ with respect to $t$ is $\frac{1}{t}$, the derivative of $\ln(t + 7)$ with respect to $t$ is $\frac{1}{t + 7}$, and the derivative of $\ln(t + 5)$ with respect to $t$ is $\frac{1}{t + 5}$. So, $\frac{1}{y}\cdot y'=\frac{1}{t}+\frac{1}{t + 7}+\frac{1}{t + 5}$.

Step3: Solve for $y'$

Multiply both sides of the equation $\frac{1}{y}\cdot y'=\frac{1}{t}+\frac{1}{t + 7}+\frac{1}{t + 5}$ by $y$. Since $y=t(t + 7)(t + 5)$, we have $y'=t(t + 7)(t + 5)\left(\frac{1}{t}+\frac{1}{t + 7}+\frac{1}{t + 5}\right)$. Expand the right - hand side: [ \begin{align*} y'&=(t + 7)(t + 5)+t(t + 5)+t(t + 7)\ &=t^{2}+12t + 35+t^{2}+5t+t^{2}+7t\ &=3t^{2}+24t + 35 \end{align*} ]

Answer:

$y'=3t^{2}+24t + 35$