Q.Find the derivative of $99x$ at $x = 100$.

We are asked to compute

1. Recognize the Riemann sum structure

The factor $\frac{1}{n}\sum_{r=1}^{n} f(r/n)$ is a standard right‑endpoint Riemann sum for $\int_0^1 f(x)\,dx$.

Here we have an extra factor of $2$, so

Thus the problem reduces to evaluating $2\int_0^1 \frac{x^3}{1-3x+3x^2}\,dx$.

2. Simplify the integrand

The denominator $1-3x+3x^2$ looks like part of $(1-x)^3$ expansion:

So

Hence

This symmetric form suggests a clever trick.

3. Use the substitution $x \mapsto 1-x$

Consider the integral $I = \int_0^1 f(x)\,dx$.

Let $t = 1-x$, then $dx = -dt$ and when $x=0$, $t=1$; when $x=1$, $t=0$. So

Renaming the dummy variable back to $x$, we have

4. Add the two expressions for $I$

Thus $2I = 1$, so $I = \frac12$.

5. Apply the factor of 2 from the original limit

The limit is $2I = 2 \cdot \frac12 = 1$.

Let $u=\dfrac{1-x^2}{1+x^2}$ and $v=\dfrac{2x}{1+x^2}$; we need $\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}$.

Differentiate $u$ (quotient rule):

Differentiate $v$:

Divide:

At $x=2$:

We have $y = \frac{ax + b}{cx + d}$. This is a rational function where $x$ is the independent variable and $y$ is expressed in terms of $x$. The question asks for $\frac{dx}{dy}$, which is the derivative of $x$ with respect to $y$. A direct approach would be to solve for $x$ in terms of $y$ and then differentiate, but that can be messy. Instead, we use a cleaner method: find $\frac{dy}{dx}$ first, then take its reciprocal.

The logic is simple: if $y$ is a function of $x$, then $\frac{dx}{dy} = \frac{1}{dy/dx}$, provided $\frac{dy}{dx} \neq 0$. This works because the derivative of the inverse function is the reciprocal of the derivative of the original function. So we compute $\frac{dy}{dx}$ using the quotient rule, then flip it.

1. Differentiate $y$ with respect to $x$ Using the quotient rule: if $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$. Here $u = ax + b$ and $v = cx + d$, so $u' = a$ and $v' = c$. Therefore:

Notice the $acx$ terms cancel neatly. The numerator $ad - bc$ is a constant — it's the determinant of the coefficients, and it determines whether the function is invertible.

2. Take the reciprocal to get $\frac{dx}{dy}$ Since $\frac{dx}{dy} = \frac{1}{dy/dx}$, we have:

But this expression is in terms of $x$, and the answer choices are in terms of $y$. We need to eliminate $x$ and express everything in $y$.

3. Express $cx + d$ in terms of $y$ From the original equation $y = \frac{ax + b}{cx + d}$, cross-multiply:

Bring terms involving $x$ together:

Solve for $x$:

Now we need $cx + d$. Substitute this $x$:

Expand the numerator:

So:

Notice $bc - ad = -(ad - bc)$.

4. Substitute back into $\frac{dx}{dy}$ We have $\frac{dx}{dy} = \frac{(cx + d)^2}{ad - bc}$. Replace $cx + d$:

Cancel one factor of $(ad - bc)$:

But $cy - a = -(a - cy)$, and squaring removes the sign: $(cy - a)^2 = (a - cy)^2$. So:

The key here is to recognise that the functional equation is not just any equation — it tells us how $f$ behaves under addition. The term $-3xy$ is a symmetric bilinear form, which strongly suggests $f$ contains a quadratic part. The given limit $\lim_{h\to 0} f(h)/h = 7$ is simply $f'(0)$, because by definition $f'(0) = \lim_{h\to 0} (f(h)-f(0))/h$, and we can find $f(0)$ from the functional equation itself.

Let’s work through it cleanly.

1. Find $f(0)$. Put $x = y = 0$ in the given equation:

Hence $f(0) = 0$.

2. Interpret the given limit. The limit $\displaystyle \lim_{h\to 0} \frac{f(h)}{h} = 7$ is exactly $f'(0)$, because $f(0)=0$:

3. Set up the derivative at a general $x$. By definition,

Use the functional equation with $y = h$:

Substitute this into the difference quotient:

4. Separate the limit. Since the limit of a sum is the sum of limits (provided each exists),

The first limit is $f'(0) = 7$, so

The problem asks us to find the second derivative of the function $y = \frac{1}{x} + \cos 2x$ with respect to $x$. This involves applying the rules of differentiation twice. We will first find the first derivative, $\frac{dy}{dx}$, and then differentiate that result to find the second derivative, $\frac{d^2y}{dx^2}$. Finally, we will manipulate the expression for $\frac{d^2y}{dx^2}$ to express it in terms of $y$ and $x$ to match one of the given options.

The key concepts involved are:

• Power Rule: For a term of the form $x^n$, its derivative is $nx^{n-1}$. This is useful for $\frac{1}{x} = x^{-1}$.
• Chain Rule: For a composite function $f(g(x))$, its derivative is $f'(g(x)) \cdot g'(x)$. This is essential for differentiating $\cos 2x$ and $\sin 2x$.
• Derivatives of Trigonometric Functions: Recall that $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\sin x) = \cos x$.

Let's proceed with the differentiation.

1. Rewrite the function for easier differentiation:

   The given function is $y = \frac{1}{x} + \cos 2x$.

   It's often easier to differentiate terms like $\frac{1}{x}$ by writing them as powers of $x$.

   So, $y = x^{-1} + \cos 2x$.

2. Find the first derivative, $\frac{dy}{dx}$:

   We differentiate each term separately.

   • For $x^{-1}$, we use the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$: $\frac{d}{dx}(x^{-1}) = (-1)x^{-1-1} = -x^{-2} = -\frac{1}{x^2}$.
   • For $\cos 2x$, we use the chain rule. Let $u = 2x$. Then $\frac{du}{dx} = 2$. $\frac{d}{dx}(\cos u) = -\sin u \cdot \frac{du}{dx} = -\sin(2x) \cdot 2 = -2\sin 2x$. Combining these, the first derivative is:

3. Find the second derivative, $\frac{d^2y}{dx^2}$: Now we differentiate $\frac{dy}{dx}$ with respect to $x$.

Again, we differentiate each term:
*   For $-\frac{1}{x^2}$, which is $-x^{-2}$, we use the power rule:
    $\frac{d}{dx}(-x^{-2}) = -(-2)x^{-2-1} = 2x^{-3} = \frac{2}{x^3}$.
*   For $-2\sin 2x$, we use the constant multiple rule and the chain rule. Let $u = 2x$. Then $\frac{du}{dx} = 2$.
    $\frac{d}{dx}(-2\sin 2x) = -2 \cdot \frac{d}{dx}(\sin 2x) = -2 \cdot (\cos u \cdot \frac{du}{dx}) = -2 \cdot (\cos(2x) \cdot 2) = -4\cos 2x$.
Combining these, the second derivative is:

4. Express $\frac{d^2y}{dx^2}$ in terms of $y$ and $x$: The options provided involve $y$. We need to substitute the original expression for $y$ back into our result for $\frac{d^2y}{dx^2}$. Recall the original function: $y = \frac{1}{x} + \cos 2x$. From this, we can express $\cos 2x$ as:

Now, substitute this into the expression for $\frac{d^2y}{dx^2}$:

Distribute the $-4$:

Rearranging the terms to match the options:

Comparing this with the given options:

(A) $\dfrac{2}{x^3} + 4y - \dfrac{4}{x}$

(B) $4y - \dfrac{4}{x} - \dfrac{2}{x^3}$

(C) $\dfrac{2}{x^3} + \dfrac{4}{x} - 4y$

(D) $4y - \dfrac{4}{x} - \dfrac{1}{x^3}$

Our result matches option (C).