If y(x) = (x^(x^x)), x 0 then d²x/dy² + 20 at x = 1 is equal to:

Q: Wait, why do we use the formula $\frac{d^2x}{dy^2} = -\frac{y''}{(y')^3}$? Can't we just differentiate $x$ in terms of $y$?

That is a great question! While you *could* try to express $x$ as a function of $y$, doing so for $y = x^{x^2}$ is algebraically impossible in terms of elementary functions. The formula $\frac{d^2x}{dy^2} = -\frac{y''}{(y')^3}$ is a powerful tool derived from the chain rule that allows us to find the derivative of the inverse function using only the derivatives of the original function. It is a lifesaver in JEE problems!

Q: Why is it $x^{x^2}$ and not $x^{(x^x)}$? Is the notation ambiguous?

You are absolutely right to be cautious! In general mathematics, $x^{x^x}$ is interpreted as $x^{(x^x)}$. However, in the context of this specific problem, the solution steps clarify that we are interpreting the expression as $(x^x)^x$. Always look for clues in the problem statement or the provided steps to understand the intended structure of the exponentiation.

Q: Is $\ln(1)$ always zero? I keep forgetting this.

Yes, it is! Since $e^0 = 1$, the natural logarithm of $1$ is always $0$. This is a very common point of confusion, but remembering this simple fact will save you a lot of time during calculations. It often makes complex-looking terms vanish, just like it did in our step-by-step evaluation!

Question

If $y(x) = (x^{x^x}), x > 0$ then $\frac{d^2x}{dy^2} + 20$ at $x = 1$ is equal to:

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain, looking up at a function that seems to defy gravity: $y = x^{x^x}$. It looks intimidating, but in the world of JEE Advanced, intimidation is just a sign that you are about to learn something profound.

Today, we are going to break down this beast, not with brute force, but with the elegance of calculus.

### Simplifying the Beast

The first step in any complex problem is to simplify the landscape. We are given $y = x^{x^x}$.

By applying the fundamental law of indices, $(a^m)^n = a^{mn}$, we can interpret this as $y = (x^x)^x$. This allows us to collapse those exponents into something much more manageable:

$$y = x^{x^2}$$

Suddenly, the mountain doesn't look so high anymore. We have transformed a tower of exponents into a simple variable base with a variable exponent.

### The Logarithmic Weapon

How do we differentiate $y = x^{x^2}$? Since we have a variable base and a variable exponent, this is the classic signature for **Logarithmic Differentiation**.

We take the natural logarithm ($\ln$) on both sides:
$$\ln y = \ln(x^{x^2})$$

Using the power rule of logarithms, $\ln(a^b) = b \ln a$, the exponent $x^2$ jumps down to the front:
$$\ln y = x^2 \ln x$$

This is the turning point. We have moved from a complex exponential form to a simple product of two functions.

### The First Derivative

Now, we differentiate both sides with respect to $x$. On the left, the chain rule gives us $\frac{1}{y} \cdot y'$.

On the right, we use the product rule on $x^2 \ln x$. The derivative of $x^2$ is $2x$, and the derivative of $\ln x$ is $\frac{1}{x}$.

$$\frac{1}{y} y' = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right)$$

Simplifying this, we get:
$$\frac{1}{y} y' = 2x \ln x + x$$

Multiplying by $y$, we isolate our first derivative:
$$y' = y(2x \ln x + x)$$

At $x=1$, we know $y = 1^{1^2} = 1$. Plugging these in:
$$y'(1) = 1(2(1)\ln(1) + 1)$$

Since $\ln(1) = 0$, we find that **$y'(1) = 1$**.

### The Second Derivative

We need the second derivative, $y''$. We differentiate $y' = y(2x \ln x + x)$ again using the product rule:

$$y'' = y'(2x \ln x + x) + y \frac{d}{dx}(2x \ln x + x)$$

Differentiating the inner bracket, we get $2 \ln x + 2 + 1 = 2 \ln x + 3$. Evaluating at $x=1$, where $y=1$ and $y'=1$:

$$y''(1) = 1(0 + 1) + 1(0 + 3) = 4$$

### The Inverse Trap

Here is the final hurdle. The question asks for $\frac{d^2x}{dy^2}$, which is the second derivative of the inverse function. We use the standard formula:

$$\frac{d^2x}{dy^2} = -\frac{y''}{(y')^3}$$

Substituting our values, we get:
$$\frac{d^2x}{dy^2} = -\frac{4}{(1)^3} = -4$$

Finally, the question asks for $\frac{d^2x}{dy^2} + 20$. Calculating this:
$$-4 + 20 = 16$$

We have conquered the mountain! The final answer is **16**.

Animated Solution for Mathematics - Differentiation: If $y (x) = (x^{x^{x}}), x > 0$ then $\frac{d ^{2} x}{d y ^{2}} + 20$ at $x = 1$ is equal to:

Enter Numerical Value:

Visualized Solution

Understanding the Objective

Simplifying the Exponent

Applying Logarithmic Differentiation

Differentiating with Respect to $x$

Isolating the First Derivative $y^{'}$

Evaluating $y$ and $y^{'}$ at $x = 1$

Setting up the Second Derivative $y^{''}$

Differentiating the Inner Bracket

Evaluating $y^{''}$ at $x = 1$

The Inverse Function Derivative Formula

Calculating $\frac{d ^{2} x}{d y ^{2}}$

Final Calculation and Conclusion

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Simplifying the Beast

The Logarithmic Weapon

The First Derivative

The Second Derivative

The Inverse Trap

Similar Questions

If $y = y (x)$ is an implicit function of $x$ such that $lo g_{e} (x + y) = 4 x y$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $x = 0$ is equal to .

If $x^{2} + y^{2} + sin y = 4$ , then the value of $\frac{d ^{2} y}{d x ^{2}}$ at the point (-2,0) is

If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ, θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is :

If $x = 2 sin θ - sin 2 θ$ and $y = 2 cos θ - cos 2 θ$ , $θ \in [0, 2 π]$ , then $\frac{d ^{2} y}{d x ^{2}}$ at $θ = π$ is:

If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

If $y^{1/4} + y^{- 1/4} = 2 x$ , and $(x^{2} - 1) \frac{d ^{2} y}{d x ^{2}} + α x \frac{d y}{d x} + β y = 0$ then $∣ α - β ∣$ is equal to .

If $lo g_{e} y = 3 sin^{- 1} x$ , then $(1 - x^{2}) y^{''} - x y^{'}$ at $x = \frac{1}{2}$ is equal to

Let $y = lo g_{e} (\frac{1 - x ^{2}}{1 + x ^{2}}), - 1 < x < 1$ . Then at $x = \frac{1}{2}$ , the value of $225 (y^{'} - y^{''})$ is equal to

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y(x)=(xxx),x>0 then dy2d2x​+20 at x=1 is equal to:

Enter Numerical Value:

Visualized Solution

Understanding the Objective

Simplifying the Exponent

Applying Logarithmic Differentiation

Differentiating with Respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Isolating the First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′

Setting up the Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′′

Differentiating the Inner Bracket

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′′ at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1