If f(x) = sin⁻¹(2·3ˣ / (1+9ˣ)), then f'(-1/2) equals :- (A) √3 log_e √3 (B) -√3 log_e 3 (C) -√3 log_

Q: Wait, why can't we just differentiate the original function $\sin^{-1}(\frac{2 \cdot 3^x}{1+9^x})$ directly using the chain rule?

You absolutely could, but it would be a nightmare of algebraic complexity! The derivative of $\sin^{-1}(u)$ is $\frac{1}{\sqrt{1-u^2}} \cdot u'$. Differentiating that fraction would lead to a massive expression that is prone to calculation errors. By using the trigonometric substitution first, we simplify the function to $f(x) = 2 \tan^{-1}(3^x)$, which is much easier to differentiate. In JEE, recognizing the structure is often the key to saving time and avoiding mistakes.

Q: How do I know if $\sin^{-1}(\sin 2\theta) = 2\theta$ is valid at $x = -\frac{1}{2}$?

Great question! This is a classic JEE trap. The identity $\sin^{-1}(\sin \alpha) = \alpha$ is only true if $\alpha$ is in the principal domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$. At $x = -\frac{1}{2}$, we have $3^x = 3^{-1/2} = \frac{1}{\sqrt{3}}$. Since $\tan \theta = 3^x$, we have $\tan \theta = \frac{1}{\sqrt{3}}$, which means $\theta = \frac{\pi}{6}$. Thus, $2\theta = \frac{\pi}{3}$. Since $\frac{\pi}{3}$ is indeed within $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the simplification $f(x) = 2\theta$ is perfectly valid.

Q: Why did we use $\tan \theta$ and not $\sin \theta$ for the substitution?

We chose $\tan \theta$ because the expression $\frac{2t}{1+t^2}$ is the standard form for $\sin 2\theta$ when $t = \tan \theta$. If we had used $\sin \theta$, we would get $\frac{2 \sin \theta}{1 + \sin^2 \theta}$, which doesn't simplify into a standard inverse trigonometric identity. Always look for the structure $\frac{2t}{1+t^2}$ as a signal to use the tangent substitution.

Question

If $f(x) = \sin^{-1} \left( \frac{2 \cdot 3^x}{1+9^x} \right)$, then $f'(-\frac{1}{2})$ equals :-

Options:

(A) $\sqrt{3} \log_e \sqrt{3}$

(B) $-\sqrt{3} \log_e 3$

(C) $-\sqrt{3} \log_e \sqrt{3}$

(D) $\sqrt{3} \log_e 3$

Sigma Percentile · Accepted Answer

### The Hidden Beauty of Inverse Trigonometry

When you first look at the function $f(x) = \sin^{-1} \left( \frac{2 \cdot 3^x}{1+9^x} ight)$, it is natural to feel a bit overwhelmed. It looks like a complex mix of inverse trigonometric functions and exponential growth.

But in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

### Phase 1: The Pattern Recognition

The key to solving this problem lies in recognizing a familiar face hidden in the algebra. Look closely at the term inside the bracket: $\frac{2 \cdot 3^x}{1+9^x}$.

If we let $t = 3^x$, then $9^x$ becomes $(3^x)^2$, or $t^2$. The expression transforms into:

$$ \frac{2t}{1+t^2} $$

Does this ring a bell? It should! This is the exact structure of the double-angle identity for sine:

$$ \sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta} $$

By identifying this, we have already won half the battle.

### Phase 2: The Substitution Strategy

Now that we see the pattern, let us make a smart move. We set $3^x = 	an 	heta$.

This substitution is the bridge between the algebraic world and the trigonometric world. Our function $f(x)$ now becomes:

$$ f(x) = \sin^{-1} \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight) $$

Using our identity, this simplifies beautifully to:

$$ f(x) = \sin^{-1}(\sin 2	heta) $$

### Phase 3: The Domain Danger Zone

Here is where many students stumble. You might be tempted to immediately cancel the $\sin^{-1}$ and $\sin$ to get $2	heta$.

But remember, the property $\sin^{-1}(\sin \alpha) = \alpha$ is only valid if $\alpha$ lies within the principal domain $[-\frac{\pi}{2}, \frac{\pi}{2}]$. We must check this for our specific point, $x = -\frac{1}{2}$.

At $x = -\frac{1}{2}$, $3^x = 3^{-1/2} = \frac{1}{\sqrt{3}}$. Since $	an 	heta = \frac{1}{\sqrt{3}}$, we find $	heta = \frac{\pi}{6}$.

Therefore, $2	heta = \frac{\pi}{3}$. Since $\frac{\pi}{3}$ is well within the safe zone of $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we are safe to proceed with $f(x) = 2	heta$.

### Phase 4: The Calculus

Now we return to the world of $x$. Since $	an 	heta = 3^x$, we have $	heta = 	an^{-1}(3^x)$.

Thus, our function is $f(x) = 2 	an^{-1}(3^x)$. This is a much friendlier function to differentiate!

Applying the chain rule, we get:

$$ f'(x) = 2 \cdot \frac{1}{1+(3^x)^2} \cdot \frac{d}{dx}(3^x) $$

Recall that the derivative of $a^x$ is $a^x \ln a$. So:

$$ f'(x) = \frac{2 \cdot 3^x \ln 3}{1+9^x} $$

### Phase 5: The Final Polish

Finally, we substitute $x = -\frac{1}{2}$ into our derivative:

$$ f'(-\frac{1}{2}) = \frac{2 \cdot 3^{-1/2} \ln 3}{1 + 9^{-1/2}} $$

Simplifying the fractions, $9^{-1/2} = \frac{1}{3}$ and $3^{-1/2} = \frac{1}{\sqrt{3}}$. The denominator becomes $1 + \frac{1}{3} = \frac{4}{3}$.

So, the calculation proceeds as follows:

$$ f'(-\frac{1}{2}) = \frac{2 \cdot (1/\sqrt{3}) \ln 3}{4/3} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} \ln 3 = \frac{\sqrt{3}}{2} \ln 3 $$

To match the standard form, we use the property $b \ln a = \ln(a^b)$, giving us the final result:

**$$ \frac{\sqrt{3}}{2} \ln 3 $$**

Animated Solution for Mathematics - Inverse Trigonometric Functions: If $f (x) = sin^{- 1} (\frac{2 \cdot 3 ^{x}}{1 + 9 ^{x}})$ , then $f^{'} (- \frac{1}{2})$ equals :-

Select Answer:

Visualized Solution

The Function Structure

Recognizing the Pattern

The Substitution

Trigonometric Transformation

Applying the Identity

The JEE Trap: Branch Analysis

Evaluating $θ$ at $x = - \frac{1}{2}$

Rewriting the Function

Differentiation Setup

Applying the Chain Rule

The Derivative Expression

Substituting $x = - \frac{1}{2}$

Simplifying the Terms

Final Calculation

Matching the Options

The Sigma Insight: Properties of Inverse Trigonometric Functions

The Hidden Beauty of Inverse Trigonometry

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=sin−1(1+9x2⋅3x​), then f′(−21​) equals :-