Derivative of $y = \log_{\sqrt[3]{x}}(7)$.

Question

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 $$\frac{dy}{dx}= \dfrac{-3\ln(7)}{x(\ln(x))^2}.$$

Answer 1

Yes, you are right. Simplify as follows $$y=\log_{\sqrt[3]{x}}(7)=\frac{\ln 7}{\ln (\sqrt[3]{x})}=\frac{\ln (7)}{\frac13\ln x}=\frac{3\ln (7)}{\ln x}$$ $$\therefore \frac{dy}{dx}=3\ln (7)\left(\frac{-1}{(\ln x)^2}\frac1x\right)=-\frac{3\ln (7)}{x(\ln x)^2}$$

Answer 2

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.

Answer 3

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\log_{\sqrt[3]{x}}(7)\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\dfrac{3\ln\left(7\right)}{\ln\left(x\right)}\right]$$

$$=3\ln\left(7\right)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{\ln\left(x\right)}\right]$$

$$=-\frac{3\ln\left(7\right)\cdot\dfrac{1}{x}}{\ln^2\left(x\right)}$$

$$=-\frac{3\ln\left(7\right)}{x\ln^2\left(x\right)}$$

Answer 4

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 :)