Q.A function $f : \mathbb{R} \to \mathbb{R}$ satisfies the equation $f(x + y) = f(x)\,f(y)$ for all $x, y \in \mathbb{R}$, $f(x) \ne 0$. Suppose that the function is differentiable at $x = 0$ and $f'(0) = 2$. Prove that $f'(x) = 2f(x)$.

Let $N(x)=e^{2x}-e^{-2x}$ and $D(x)=e^{3x}+e^{-3x}$, so $f=N/D$.

$D(0)=2,\quad D'(x)=3e^{3x}-3e^{-3x}\implies D'(0)=0.$ …

We are told $f$ is a linear polynomial, so $f(x)=mx+c$ for constants $m$ and $c$. The functional equation $f(ax+by)=af(x)+bf(y)$ must hold for all real $x,y$. This is a strong condition — it essentially says $f$ is both additive and homogeneous in a very specific way. The trick is to realize that such an equation forces $a+b=1$ (otherwise the left and right sides can't match for all $x,y$), and then the rest is just plugging in.

Let’s work through it step by step.

1. Write the general form of $f$. Since $f$ is linear, $f(x)=mx+c$. The derivative $f'(0)=5$ tells us $m=5$. The value $f(0)=7$ gives $c=7$. So

2. Plug into the functional equation. The left side:

The right side:

3. Equate for all $x,y$. For these to be equal for every real $x$ and $y$, the coefficients of $x$ and $y$ already match ($5a$ and $5b$ on both sides). The constant terms must also match:

So the functional equation forces $a+b=1$. …

We start with $f(x) = e^x$. The composition $(f \circ f)(x)$ means $f(f(x))$: first apply $f$ to $x$, getting $e^x$, then apply $f$ again to that result. So

Now we need $\dfrac{h'(x)}{h(x)}$. This expression is the logarithmic derivative of $h$ — it’s exactly the derivative of $\log h(x)$. That’s a useful insight: if we can find $\log h(x)$, we can differentiate it directly.

1. Find $\log h(x)$ Since $h(x) = e^{e^x}$, taking the natural log gives

So $\log h(x)$ is simply $e^x$.

2. Differentiate $\log h(x)$ The derivative of $\log h(x)$ with respect to $x$ is $\dfrac{h'(x)}{h(x)}$ by the chain rule. Differentiating $e^x$ gives $e^x$. Therefore

3. Match to the options …

We are told that $g(x)$ is an antiderivative of $f(x)$, meaning $g'(x) = f(x)$. The expression $\log_e(1+(g(x))^2) + c$ is given as an antiderivative of some function — call it $h(x)$. That means the derivative of this expression with respect to $x$ equals $h(x)$.

So the problem reduces to: differentiate $\log_e(1+(g(x))^2) + c$, simplify using $g'(x)=f(x)$, and see which option matches.

1. Differentiate the given antiderivative. The derivative of $\log_e(1+(g(x))^2)$ is, by the chain rule,

Since $g'(x) = f(x)$, this becomes

2. The constant $c$ differentiates to zero, so the derivative — and hence the function for which the given expression is an antiderivative — is exactly

3. Compare with the options. …

Step-by-Step

For $\sqrt{g(x)}$, $\dfrac{d}{dx}\sqrt{g(x)} = \dfrac{g'(x)}{2\sqrt{g(x)}}$.

1. First term $\sqrt{\log(x^2+x+1)}$.

At $x=0$: $x^2+x+1 = 1$ and $\log 1 = 0$, so this is $\dfrac{1}{2\cdot 1\cdot\sqrt{0}}$ — undefined; the derivative blows up.

2. Second term $\sqrt{\cosh(2x-3)}$.

$\frac{d}{dx}\sqrt{\cosh(2x-3)} = \frac{2\sinh(2x-3)}{2\sqrt{\cosh(2x-3)}} = \frac{\sinh(2x-3)}{\sqrt{\cosh(2x-3)}}.$ …

We are told that

This means $f(x)$ is exactly the logarithmic derivative of the function $g(x) = \frac{x-1}{x-\sqrt{x}} e^{2x+1}$. That is, if $g'(x) = g(x) f(x)$, then $f(x) = \frac{g'(x)}{g(x)} = \frac{d}{dx} \log g(x)$. So instead of doing a messy product/quotient differentiation directly, we can compute $\log g(x)$, differentiate, and simplify — that’s the clean path.

1. Simplify the function first

Notice $x - \sqrt{x} = \sqrt{x}(\sqrt{x} - 1)$. So

Thus

2. Take the natural log

3. Differentiate

Simplify the fraction: multiply numerator and denominator by $x^{3/2}$:

But better: write $1 + x^{-1/2} = \frac{\sqrt{x}+1}{\sqrt{x}}$, so

The problem gives you two differentiable functions where one is the inverse of the other: $g = f^{-1}$. That means $f(g(x)) = x$ for all $x$ in the domain. When you have an inverse relationship, derivatives are linked by a simple reciprocal relation — but you have to be careful about which point you evaluate at.

The core concept: if $f$ and $g$ are inverses, then differentiating $f(g(x)) = x$ using the chain rule gives $f'(g(x)) \cdot g'(x) = 1$. So $f'(g(x)) = \frac{1}{g'(x)}$, provided $g'(x) \neq 0$. This is the inverse function theorem in its simplest form.

Now we apply it step by step.

1. We know $g(5) = 6$. That means $f(6) = 5$, because $g = f^{-1}$ — applying $f$ to both sides of $g(5)=6$ gives $f(6)=5$.

2. We also know $g'(5) = \frac{3}{4}$.

3. From the inverse relation, at $x = 5$:

Substituting $g(5)=6$ and $g'(5)=\frac{3}{4}$:

4. Solve for $f'(6)$: $f'(6) = \frac{1}{3/4} = \frac{4}{3}$ …

When you see a function of the form $[g(x)]^{h(x)}$ — where both the base and the exponent are functions of $x$ — the standard power rule or exponential rule alone won't work. The cleanest method is to take the natural logarithm of both sides, differentiate implicitly, and then solve for $f'(x)$. This is called logarithmic differentiation, and it turns the messy exponent into a product you can handle with the product rule.

Here, $f(x) = (\log x)^{\sin x}$ with $x > e$ (so $\log x > 1$, and the function is well-defined and positive). We want $f'(\pi)$.

1. Take logs. Let $y = f(x) = (\log x)^{\sin x}$. Then

The right side is now a product of two functions of $x$, which is much easier to differentiate.

2. Differentiate both sides with respect to $x$. On the left, by the chain rule: $\frac{d}{dx}(\log y) = \frac{1}{y} \cdot y' = \frac{y'}{y}$. On the right, use the product rule:

(The derivative of $\log(\log x)$ is $\frac{1}{\log x} \cdot \frac{1}{x}$ by the chain rule.)

So we have:

3. Solve for $y'$. Multiply through by $y = (\log x)^{\sin x}$:

4. Evaluate at $x = \pi$. At $x = \pi$, we have $\sin \pi = 0$ and $\cos \pi = -1$. …

Concept — logarithmic differentiation. When the variable appears in both base and exponent, take natural logs first.

Step 1 — take logarithms.

Step 2 — differentiate implicitly w.r.t. $x$.

Step 3 — find $y$ at $x=1$. The original relation at $x=1$: $(3y)^2 = 5\cdot 2^3 = 40$, so $3y = \sqrt{40} = 2\sqrt{10}$ (positive root for the log to exist), i.e. $y = \dfrac{2\sqrt{10}}{3}$, and $\log(3y) = \tfrac{1}{2}\log 40$.

Step 4 — substitute $x=1$. …

The key idea: $|x-a|$ has derivative $+1$ for $x > a$ and $-1$ for $x < a$ (it's undefined at $x=a$). So $f(x) = |x-1| + |x-2|$ is a sum of two such V-shaped functions. Its derivative is simply the sum of the derivatives of each piece, as long as we avoid the points $x=1$ and $x=2$ where the absolute values have corners.

We just need to figure out, for each given $x$, whether it lies to the left or right of $1$ and $2$, then add the corresponding $\pm 1$ contributions.

1. For $x = -2023$

   This is far to the left of both $1$ and $2$.

   • For $|x-1|$: since $x < 1$, derivative is $-1$.
   • For $|x-2|$: since $x < 2$, derivative is $-1$. So $f'(-2023) = (-1) + (-1) = -2$.

2. For $x = \dfrac{2024}{2023}$

   Note that $\dfrac{2024}{2023} \approx 1.0005$, so it's just slightly greater than $1$ but still less than $2$.

   • For $|x-1|$: since $x > 1$, derivative is $+1$.
   • For $|x-2|$: since $x < 2$, derivative is $-1$. So $f'\left(\dfrac{2024}{2023}\right) = (+1) + (-1) = 0$.

3. For $x = 2023$

   This is far to the right of both $1$ and $2$.

   • For $|x-1|$: since $x > 1$, derivative is $+1$. …