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

Concept: Chain Rule — differentiate the outer function, then multiply by the derivative of the inner function.

Let $y = \sin(ax + b)$.

The outer function is $\sin u$, whose derivative is $\cos u$.

The inner function is $u = ax + b$, whose derivative is $a$.

Applying the chain rule:

The problem asks for the derivative of $\sin(ax+b)$ with respect to $x$. This is a classic composition of two functions: an outer sine function and an inner linear function $ax+b$. The Chain Rule is the natural tool here — it tells us to differentiate the outer function first, leaving the inner untouched, then multiply by the derivative of the inner function.

Let’s walk through it step by step.

1. Identify the composition.

   We have $y = \sin(u)$ where $u = ax + b$.

   The outer function is $\sin(u)$, and the inner function is $u = ax + b$.

2. Differentiate the outer function with respect to its argument.

   The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$.

   So, $\frac{dy}{du} = \cos(u)$.

3. Differentiate the inner function with respect to $x$.

   The derivative of $ax + b$ with respect to $x$ is simply $a$ (since $b$ is constant).

   So, $\frac{du}{dx} = a$.

4. Apply the Chain Rule.

   The Chain Rule states:

Substituting what we have:

Method: Differentiating a Trig Function of a Linear Expression

This is the simplest chain-rule case, and it produces a reusable pattern worth memorising: the derivative of $\sin(kx+c)$ (or $\cos$, $\tan$, etc.) is always the outer derivative times the constant $k$.

Steps

Step 1: Identify the inner linear expression

Write the argument of the trig function as $u = (\text{coefficient})\cdot x + (\text{constant})$.

Step 2: Differentiate the outer trig function with respect to $u$

Apply the standard rule (e.g. $\frac{d}{du}\sin u = \cos u$), keeping the result in terms of $u$.

Step 3: Differentiate the inner linear expression with respect to $x$

The derivative of $(\text{coefficient})\cdot x + (\text{constant})$ is simply the coefficient — the additive constant contributes nothing, since its derivative is zero.

Step 4: Multiply and substitute back

Step 5: Recognise the reusable pattern

For any $\sin(kx+c)$, the derivative is always $k\cos(kx+c)$ — the same pattern extends directly to $\cos(kx+c) \to -k\sin(kx+c)$ and other trig functions, so this shortcut is worth remembering rather than re-deriving each time.

Common Mistakes

Mistake 1: Forgetting to multiply by the coefficient $a$

Why it's wrong: writing $\frac{dy}{dx} = \cos(ax+b)$ without the leading factor of $a$ ignores the inner derivative of the linear expression $ax+b$, which is $a$ (not $1$). Correct approach: always compute the inner derivative separately — even when it looks like a trivial linear expression, its derivative (the coefficient) must still be multiplied into the final answer.

Mistake 2: Treating the constant $b$ as if it contributes to the derivative

Why it's wrong: some students mistakenly think a nonzero constant $b$ should appear somewhere in the final derivative — but the derivative of any additive constant is always zero, so $b$ only affects the argument of the cosine, never the multiplying factor out front. Correct approach: remember that only the coefficient of $x$ (here, $a$) survives differentiation of a linear inner function; any purely additive constant disappears completely.