Find y' for y = log_2(1-x cos x)
Find $y'$ for $y = \log_{2}(1-x\cos x)$.
Explanation of Method
A base-2 logarithm with a product inside.
Step 1: Recall the rule for a general base
From Problem 19, $\bigl(\log_{a} u\bigr)' = \dfrac{u'}{u\ln a}$. Here $a = 2$ and $u = 1-x\cos x$.
Step 2: Differentiate the inner function
The product $x\cos x$ needs the product rule, and the whole thing carries a minus sign: \[u' = -\bigl(\cos x + x(-\sin x)\bigr) = -\cos x + x\sin x = x\sin x - \cos x.\]
Step 3: Combine
\[ y' = \frac{x\sin x - \cos x}{\bigl(1-x\cos x\bigr)\ln 2}. \]
Step 4: Verification
\[x = 0:\quad u = 1 \;\Longrightarrow\; y = 0,\quad y' = \frac{0-1}{1\cdot\ln 2} = -\frac{1}{\ln 2} \approx -1.44270\]
\[\frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx -1.44270 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{x\sin x-\cos x}{\bigl(1-x\cos x\bigr)\ln 2}\,}\]
Remark
Remark. The constant $\dfrac{1}{\ln 2} = \log_{2}e \approx 1.4427$ is the only trace the base leaves: switching to natural logarithms would remove it entirely. The domain requires $1-x\cos x \gt 0$, which holds near the origin but fails for large $x$ where $x\cos x$ exceeds $1$.