Q.Differentiate with respect to $x$: $(ax^2 + \cot x)(p + q\cos x)$.

This problem requires the application of the Product Rule for differentiation.

Let $u(x) = ax^2 + \cot x$ and $v(x) = p + q\cos x$.

1. Differentiate $u(x)$ with respect to $x$:

   $u'(x) = \frac{d}{dx}(ax^2 + \cot x) = 2ax - \csc^2 x$.

2. Differentiate $v(x)$ with respect to $x$:

   $v'(x) = \frac{d}{dx}(p + q\cos x) = 0 + q(-\sin x) = -q\sin x$.

3. Apply the Product Rule: …

When faced with differentiating a function that is a product of two other functions, like $f(x) = g(x)h(x)$, we cannot simply differentiate each part separately and multiply the results. This is a common mistake. Instead, we use the Product Rule.

The intuition behind the Product Rule comes from considering how a small change in $x$ affects the product $uv$. If $u$ changes by $\Delta u$ and $v$ changes by $\Delta v$, the new product is $(u + \Delta u)(v + \Delta v) = uv + u\Delta v + v\Delta u + \Delta u \Delta v$. The change in the product is $u\Delta v + v\Delta u + \Delta u \Delta v$. When we take the limit as $\Delta x \to 0$, the $\Delta u \Delta v$ term (which is a product of two small changes) becomes negligible compared to the other terms, leading to the Product Rule.

We will also need the derivatives of some standard functions:

• $\frac{d}{dx}(x^n) = nx^{n-1}$
• $\frac{d}{dx}(\text{constant}) = 0$
• $\frac{d}{dx}(\cot x) = -\csc^2 x$
• $\frac{d}{dx}(\cos x) = -\sin x$

Let's differentiate the given expression step-by-step.

1. Identify the two functions, $u$ and $v$.

   The given expression is $(ax^2 + \cot x)(p + q\cos x)$.

   Let $u = ax^2 + \cot x$.

   Let $v = p + q\cos x$.

   Here, $a, p, q$ are constants.

2. Calculate the derivative of $u$ with respect to $x$, denoted as $u'$.

   $u = ax^2 + \cot x$

   We differentiate each term using the sum rule and standard derivative formulas:

   $\frac{du}{dx} = \frac{d}{dx}(ax^2) + \frac{d}{dx}(\cot x)$

   $\frac{d}{dx}(ax^2) = a \cdot (2x^{2-1}) = 2ax$

   $\frac{d}{dx}(\cot x) = -\csc^2 x$

   So, $u' = 2ax - \csc^2 x$.

3. Calculate the derivative of $v$ with respect to $x$, denoted as $v'$.

   $v = p + q\cos x$

   We differentiate each term:

   $\frac{dv}{dx} = \frac{d}{dx}(p) + \frac{d}{dx}(q\cos x)$

   Since $p$ is a constant, $\frac{d}{dx}(p) = 0$.

   For $q\cos x$, $q$ is a constant multiplier: $\frac{d}{dx}(q\cos x) = q \cdot \frac{d}{dx}(\cos x) = q(-\sin x) = -q\sin x$.

   So, $v' = -q\sin x$.

4. Apply the Product Rule formula.

   Now we substitute $u, v, u', v'$ into the Product Rule formula: $\frac{dy}{dx} = u'v + uv'$.

   $\frac{dy}{dx} = (2ax - \csc^2 x)(p + q\cos x) + (ax^2 + \cot x)(-q\sin x)$ …