question\nfind the derivative of y = (7x^4 - 6)^x. be sure to include parentheses around the arguments of…

Explanation:

Step1: Take natural - log of both sides

Let $y=(7x^{4}-6)^{x}$. Then $\ln y = x\ln(7x^{4}-6)$.

Step2: Differentiate both sides with respect to $x$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v=\ln(7x^{4}-6)$. The derivative of $\ln y$ with respect to $x$ is $\frac{y^\prime}{y}$, the derivative of $x$ is $1$, and the derivative of $\ln(7x^{4}-6)$ using the chain - rule is $\frac{28x^{3}}{7x^{4}-6}$. So, $\frac{y^\prime}{y}=1\times\ln(7x^{4}-6)+x\times\frac{28x^{3}}{7x^{4}-6}$.

Step3: Solve for $y^\prime$

Multiply both sides by $y=(7x^{4}-6)^{x}$. Then $y^\prime=(7x^{4}-6)^{x}\left[\ln(7x^{4}-6)+\frac{28x^{4}}{7x^{4}-6}\right]$.

Answer:

$(7x^{4}-6)^{x}\left[\ln(7x^{4}-6)+\frac{28x^{4}}{7x^{4}-6}\right]$