Derivative of $y = \log_{\sqrt[3]{x}}(7)$. Never dealt with a derivative of these type. My approach was $$y = \log_{\sqrt[3]{x}}(7) \iff 7 = (\sqrt[3]{x})^y.$$ Then,
$$\frac{d}{dx}(7) = \frac{d}{dx}\left(\sqrt[3]{x}\right)^y \Rightarrow (\sqrt[3]{x})^y = e^{\frac{y\ln(x)}{3}} $$
From here,
$0 = e^u\dfrac{du}{dx}$ and $u = \dfrac{y\ln(x)}{3}.$ Thus,
$$0 = \frac{du}{dx} = \frac{y}{3x} +\frac{\ln(x)}{3}\frac{dy}{dx}.$$
Which implies that $$\frac{dy}{dx}= \frac{-\log_{\sqrt[3]{x}}(7)}{x\ln(x)}.$$
Is this the correct derivative? Can I alternatively use $\log_{b}(a) = \dfrac{\ln(a)}{\ln(b)}$, with $b = \sqrt[3]{x}$ and $a=7$? In that case, I arrive at
$$\frac{dy}{dx}= \dfrac{-3\ln(7)}{x(\ln(x))^2}.$$
 A: Yes, you are right. Simplify as follows
$$y=\log_{\sqrt[3]{x}}(7)=\frac{\ln 7}{\ln (\sqrt[3]{x})}=\frac{\ln (7)}{\frac13\ln x}=\frac{3\ln (7)}{\ln x}$$
$$\therefore \frac{dy}{dx}=3\ln (7)\left(\frac{-1}{(\ln x)^2}\frac1x\right)=-\frac{3\ln (7)}{x(\ln x)^2}$$
A: Note
$$ \frac{-\log_{\sqrt[3]{x}}(7)}{x\ln x}
= \frac{-\frac{\ln 7}{\frac13\ln x}}{x\ln x}
= \frac{-3\ln 7}{x(\ln x)^2}
$$
So, either method is correct; they yield the same results.
A: $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\log_{\sqrt[3]{x}}(7)\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\dfrac{3\ln\left(7\right)}{\ln\left(x\right)}\right]$$
$$=3\ln\left(7\right)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{\ln\left(x\right)}\right]$$
$$=-\frac{3\ln\left(7\right)\cdot\dfrac{1}{x}}{\ln^2\left(x\right)}$$
$$=-\frac{3\ln\left(7\right)}{x\ln^2\left(x\right)}$$
A: Let's do what any good mathematician would: Generalize! Suppose $y=\log_{f(x)}(g(x))$. Then we can use the change-of-base formula:
$$y=\frac{\ln(g(x))}{\ln(f(x))}$$
And now use the quotient rule:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{g'(x)}{g(x)}\ln(f(x))-\frac{f'(x)}{f(x)}\ln(g(x))}{\ln(f(x))^2}$$
Plug in $f(x)=x^{1/3}$, $g(x)=7$ to find your answer :)
