Q.Find $\frac{dy}{dx}$ in the following: $\frac{\cos x}{\log x}, x > 0$

The key idea is Implicit Differentiation — but here the function is given explicitly as $y = \frac{\cos x}{\log x}$, so we simply differentiate using the quotient rule.

Let $u = \cos x$ and $v = \log x$ (natural log). Then:

We have $u' = -\sin x$ and $v' = \frac{1}{x}$. Substituting:

$\frac{dy}{dx} = \frac{(\log x)(-\sin x) - (\cos x)\left(\frac{1}{x}\right)}{(\log x)^2}$ …

The problem asks for $\frac{dy}{dx}$ of $y = \frac{\cos x}{\log x}$, with $x > 0$. The condition $x > 0$ ensures the logarithm is defined and the denominator is non-zero (except at $x=1$, but we differentiate away from that point).

The core idea here is the quotient rule. Whenever you have a function that is one differentiable function divided by another, you don't need to rewrite it or use the product rule with a negative exponent (though that also works). The quotient rule is direct and clean.

Let’s apply it step by step.

1. Identify the numerator and denominator.

   Here, $u = \cos x$ and $v = \log x$. Both are differentiable for $x > 0$.

2. Differentiate each part separately.

   • $\frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x$
   • $\frac{dv}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x}$ (Remember: $\log x$ here means the natural logarithm, as is standard in calculus.)

3. Plug into the quotient rule formula.

4. Simplify the numerator. The numerator becomes $-\sin x \log x - \frac{\cos x}{x}$. There’s no further algebraic simplification that makes it cleaner, so we leave it as is. …

Method: The Quotient Rule with a Logarithmic Denominator

Use this whenever $y$ is a ratio where the denominator is a logarithm (or another standard function whose own derivative you must recall correctly).

Steps

Step 1: Identify the numerator and denominator functions

Write $y=\dfrac{u}{v}$ with $u=\cos x$ and $v=\log x$, valid for $x>0$ (so the logarithm is defined) and excluding $x=1$ (where $\log x=0$ would make the denominator zero).

Step 2: Recall the quotient rule formula

Step 3: Differentiate $u$ and $v$ separately, using the correct standard derivative for each

$\frac{du}{dx} = -\sin x, \qquad \frac{dv}{dx} = \frac{1}{x}$ …

Common Mistakes

Mistake 1: Misremembering the derivative of $\log x$

Why it's wrong: writing $\dfrac{d}{dx}\log x = \dfrac{1}{\log x}$ instead of $\dfrac1x$ is one of the most frequent errors with logarithmic denominators — it conflates the function's value with its derivative. Correct approach: fix $\dfrac{d}{dx}\log x=\dfrac1x$ firmly and double-check it every time a logarithm appears in a quotient.

Mistake 2: Swapping the order of the two terms in the quotient rule numerator

Why it's wrong: the formula is $v\,u' - u\,v'$, and reversing it to $u\,v' - v\,u'$ flips the overall sign. Correct approach: always write "denominator times derivative of numerator, minus numerator times derivative of denominator" before substituting. …