Q.For the function $f(x) = \dfrac{x^{100}}{100} + \dfrac{x^{99}}{99} + \cdots + \dfrac{x^2}{2} + x + 1$. Prove that $f'(1) = 100\,f'(0)$.

Step 1 — Simplify $f(x)$

For $|x| \le 1$ (which includes $x = \tfrac12$):

Step 2 — Differentiate

Step 3 — Evaluate at $x = \tfrac12$ …

The question gives us a function and asks us to match its derivative to a pattern: $ma + nb - p$, where $m$, $n$, and $p$ are expressions in $x$ (and possibly constants). The key is to differentiate term-by-term, then compare coefficients of $a$ and $b$, and the remaining trigonometric term.

Notice that $a$ and $b$ are constants — they are not variables. So when we differentiate, they simply multiply the derivative of whatever they are attached to. The pattern $ma + nb - p$ means the derivative should look like:

(something) $\times a$ + (something else) $\times b$ — (a third term). That third term comes from the $\cos x$ part.

Let’s work through it.

1. Rewrite the function for clarity

   $f(x) = \dfrac{a}{x^4} - \dfrac{b}{x^2} + \cos x$

   It helps to write the first two terms with negative exponents:

   $f(x) = a x^{-4} - b x^{-2} + \cos x$

2. Differentiate term by term

   • For $a x^{-4}$: $\frac{d}{dx}(a x^{-4}) = a \cdot (-4) x^{-5} = -\dfrac{4a}{x^5}$
   • For $-b x^{-2}$: $\frac{d}{dx}(-b x^{-2}) = -b \cdot (-2) x^{-3} = \dfrac{2b}{x^3}$ (Be careful: the minus sign in front of $b$ multiplies the derivative of $x^{-2}$, which is $-2x^{-3}$, giving a positive result.)
   • For $\cos x$: $\frac{d}{dx}(\cos x) = -\sin x$

   So the derivative is:

   $f'(x) = -\dfrac{4a}{x^5} + \dfrac{2b}{x^3} - \sin x$

3. Match to the form $ma + nb - p$

   Compare:

   $-\dfrac{4a}{x^5} + \dfrac{2b}{x^3} - \sin x \quad \longleftrightarrow \quad m a + n b - p$

   This gives: …

Step 1 — Recognise the identity.

Put $x=\tan\theta$, so $\theta=\tan^{-1}x$. Then

and

Step 2 — The branch (this is the entire point of the question).

$\sin^{-1}(\sin u)=u$ only when $u\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$; if $u\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$ then $\sin^{-1}(\sin u)=\pi-u$.

Here $u=2\theta=2\tan^{-1}x$:

• If $|x|\le 1$ then $|\theta|\le\frac{\pi}{4}$, so $|2\theta|\le\frac{\pi}{2}$ and $f(x)=2\tan^{-1}x$.
• If $x>1$ then $\theta>\frac{\pi}{4}$, so $2\theta>\frac{\pi}{2}$ — out of the principal range — and

Step 3 — Which branch is $x=\sqrt3$ on?

$\sqrt3\approx1.732>1\quad\Longrightarrow\quad \text{the second branch}: \ f(x)=\pi-2\tan^{-1}x .$ …

The key insight here is that the expression inside the inverse sine looks like something we can simplify. When you see $\frac{2^{x+1}}{1+4^x}$, notice that $4^x = (2^x)^2$ and $2^{x+1} = 2 \cdot 2^x$. So the whole thing becomes $\frac{2 \cdot 2^x}{1 + (2^x)^2}$. That form — $\frac{2t}{1+t^2}$ — is a dead giveaway for the double-angle formula for tangent: $\sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2 \tan^{-1} t$ for $t$ in the right range. Here $t = 2^x$, which is always positive, so the identity holds cleanly.

Once we have $f(x) = 2 \tan^{-1}(2^x)$, differentiation is just the chain rule applied to a standard derivative. Let’s go step by step.

1. Rewrite the function Let $t = 2^x$. Then $4^x = (2^x)^2 = t^2$, and $2^{x+1} = 2 \cdot 2^x = 2t$. So

2. Apply the inverse sine identity For $t > 0$, we have the identity

(This comes from letting $\theta = \tan^{-1} t$, so $\sin 2\theta = \frac{2t}{1+t^2}$.)

Hence

3. Differentiate Differentiate $f(x) = 2 \tan^{-1}(2^x)$ using the chain rule. Recall $\frac{d}{dx} \tan^{-1} u = \frac{1}{1+u^2} \cdot \frac{du}{dx}$. Here $u = 2^x$, so $\frac{du}{dx} = 2^x \log 2$. Thus …