Find y' for y = log_x 2

Find $y'$ for $y = \log_{x} 2$.

Step 1: Read the notation carefully

Here the base is the variable and the argument is the constant $2$ — the reverse of the usual situation. There is no differentiation rule for $\log_{x}$ as it stands, so the first step is to convert to natural logarithms.

Step 2: Change of base

The change-of-base formula $\log_{b}a = \dfrac{\ln a}{\ln b}$ gives

\[ y = \log_{x}2 = \frac{\ln 2}{\ln x}. \]

The numerator is now a constant; only $\ln x$ depends on $x$. The domain is $x>0,\ x\neq 1$, since $\ln x$ must be nonzero.

Step 3: Differentiate

Write $y = \ln 2\cdot\bigl(\ln x\bigr)^{-1}$ and use the power and chain rules: $y' = \ln 2\cdot(-1)\bigl(\ln x\bigr)^{-2}\cdot\frac{1}{x} = -\frac{\ln 2}{x\ln^{2}x}$.

Step 4: Reading the answer

Since $\ln 2 > 0$, $x > 0$ and $\ln^{2}x > 0$, the derivative is negative everywhere on the domain. The function decreases on $(0,1)$ and, separately, on $(1,\infty)$ — it is not decreasing "across" $x=1$, where it has an infinite discontinuity: $y\to-\infty$ as $x\to1^{-}$ and $y\to+\infty$ as $x\to1^{+}$.

Step 5: Verification

\[x = e:\quad y = \frac{\ln 2}{1} = \ln 2, \qquad y' = -\frac{\ln 2}{e\cdot 1} \approx -0.25499\]

\[\text{Numerically: } \frac{y(e+10^{-5})-y(e-10^{-5})}{2\times10^{-5}} \approx -0.254995 \;\;\checkmark\]

\[x = 2:\quad y = \log_{2}2 = 1 \;\;\checkmark, \qquad y' = -\frac{\ln 2}{2\ln^{2}2} = -\frac{1}{2\ln 2} \approx -0.72135\]