Question

1. find $\frac{d}{dt}ln 7x^{3}$
2. find the equation of the tange
3. find the equation of the tange an exact value for the slope, do
4. evaluate $\frac{d}{dx}\frac{3xln(2x^{3}-5x + 4)}{4 - 3x}$

Explanation:

Step1: Use chain - rule and log derivative

If $y = \ln(7x^{3})$, first recall that $\ln(ab)=\ln a+\ln b$, so $\ln(7x^{3})=\ln 7+\ln(x^{3})$. And $\frac{d}{dt}(\ln 7 + \ln(x^{3}))=\frac{d}{dt}(\ln 7)+\frac{d}{dt}(\ln(x^{3}))$. Since $\ln 7$ is a constant, $\frac{d}{dt}(\ln 7) = 0$. Also, $\ln(x^{3}) = 3\ln x$. Then by the chain - rule $\frac{d}{dt}(3\ln x)=3\frac{1}{x}\frac{dx}{dt}$.

Answer:

$3\frac{1}{x}\frac{dx}{dt}$