Q.Find $\frac{dy}{dx}$ in the following: $\frac{\sin (ax + b)}{\cos (cx + d)}$

Use the quotient rule $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}$, with the chain rule on each linear argument.

Let $u=\sin(ax+b)$ and $v=\cos(cx+d)$, so $u'=a\cos(ax+b)$ and $v'=-c\sin(cx+d)$. Then …

We are differentiating $y=\dfrac{\sin(ax+b)}{\cos(cx+d)}$, a ratio of two functions, so the quotient rule is the tool. Because each trig function has a linear argument, the chain rule supplies the constants $a$ and $c$.

Set up

Let $u=\sin(ax+b)$ and $v=\cos(cx+d)$. Then

Apply the quotient rule

Simplify the numerator

The double negative becomes a plus: …

Method: Quotient Rule Combined with the Chain Rule (Linear Arguments)

Use this method whenever you must differentiate a ratio of two functions, $y = \dfrac{u(x)}{v(x)}$, and each of $u$ and $v$ is itself a function of a linear expression (like $ax+b$) rather than of plain $x$.

Steps

Step 1: Identify the numerator and denominator as separate functions

Label $u(x)$ as the numerator and $v(x)$ as the denominator before differentiating anything — do not try to simplify or combine the expression first.

Step 2: Differentiate $u$ and $v$ separately, using the chain rule for their linear arguments

Because the argument is $mx+n$ rather than plain $x$, every derivative picks up the constant multiplier $m$:

In general, $\frac{d}{dx}f(mx+n) = m\cdot f'(mx+n)$ for any linear inner function.

Step 3: Apply the quotient rule formula …

Common Mistakes

Mistake 1: Dropping the chain-rule constant $a$ or $c$

Why it's wrong: writing $\frac{d}{dx}\sin(ax+b)=\cos(ax+b)$ (missing the factor $a$) treats the argument as if it were plain $x$. Correct approach: whenever the argument of a trig function is $mx+n$ rather than $x$, the derivative always carries an extra factor of $m$ from the chain rule.

Mistake 2: Losing the sign when substituting $v'=-c\sin(cx+d)$ into the quotient rule

Why it's wrong: the quotient rule's $-uv'$ term becomes $-u\cdot(-c\sin(cx+d)) = +cu\sin(cx+d)$; students often keep the minus sign and write the numerator with the wrong sign on the second term. Correct approach: substitute $v'$ with its own sign intact and simplify the double negative as a separate, explicit step. …