Q.Find $\frac{dy}{dx}$ in the following: $\cos (a \cos x + b \sin x)$, for some constant $a$ and $b$.

The key idea is the Chain Rule: differentiate the outer function (cosine) first, then multiply by the derivative of the inner function.

Step 1: Let $u = a \cos x + b \sin x$. Then $y = \cos u$.

Step 2:

$\frac{dy}{du} = -\sin u$

$\frac{du}{dx} = -a \sin x + b \cos x$

Step 3: By the chain rule, …

The problem asks for $\frac{dy}{dx}$ when $y = \cos(a\cos x + b\sin x)$, where $a$ and $b$ are constants. This is a classic composite function — a cosine of a linear combination of $\sin x$ and $\cos x$.

The Chain Rule is the natural tool here. It says: if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. In words: differentiate the outer function, keep the inner function untouched, then multiply by the derivative of the inner function.

Let’s apply it step by step.

1. Identify the outer and inner functions.

   Outer: $f(u) = \cos u$, where $u = a\cos x + b\sin x$ is the inner function.

2. Differentiate the outer function.

   The derivative of $\cos u$ with respect to $u$ is $-\sin u$. So $f'(u) = -\sin u$.

3. Keep the inner function unchanged inside the derivative of the outer.

   That gives $-\sin(a\cos x + b\sin x)$.

4. Differentiate the inner function $u = a\cos x + b\sin x$ with respect to $x$.

   Since $a$ and $b$ are constants:

   $\frac{d}{dx}(a\cos x) = a(-\sin x) = -a\sin x$

   $\frac{d}{dx}(b\sin x) = b\cos x$

   So $u' = -a\sin x + b\cos x$.

5. Multiply the two results by the Chain Rule. …

Method: The Chain Rule for a Composite Function

Whenever a function is "wrapped inside" another function — $f(g(x))$ — differentiate the outer function first (with respect to its own argument), then multiply by the derivative of the inner function.

Steps

Step 1: Identify the outer function and the inner function

Write $y=f(u)$ where $u=g(x)$ is everything "inside" the outermost operation.

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

Use the standard derivative rule for whatever the outer function is (log, power, trig, exponential, ...), keeping $u$ untouched.

Step 3: Differentiate the inner function $u$ with respect to $x$

Step 4: Multiply the two results

$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.$ …

Common Mistakes

Mistake 1: Forgetting the minus sign from differentiating the outer $\cos u$.

Why it's wrong: $\frac{d}{du}\cos u=-\sin u$, not $+\sin u$ — dropping this sign flips the overall sign of the answer. Correct approach: always write the standard cosine derivative with its negative sign attached, never from memory alone mid-problem.

Mistake 2: Mis-differentiating $a\cos x$ as $a\sin x$ instead of $-a\sin x$. …