Problem 259 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } y=e^{\sqrt{x}}, \text { find } \frac{d^{2} y}{d x^{2}} $$

Step-by-Step Solution

Answer

The second derivative of the function \(y = e^{\sqrt{x}}\) with respect to x is \(\frac{d^2y}{dx^2} = -\frac{1}{4}e^{\sqrt{x}}x^{-3/2} + \frac{1}{4}e^{\sqrt{x}}x^{-1}\).

Step 1: Find \(\frac{dy}{du}\)

\(\frac{dy}{du} = e^u\) Next, we need to find the derivative of \(u\) with respect to x:

Step 2: Find \(\frac{du}{dx}\)

Since \(u = \sqrt{x}\) or \(u = x^{1/2}\), we have: \(\frac{du}{dx} = \frac{1}{2}x^{-1/2}\). Now we can find \(\frac{dy}{dx}\):

Step 3: Find \(\frac{dy}{dx}\)

Using the chain rule, we have: \(\frac{dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} = e^u\cdot\frac{1}{2}x^{-1/2} = e^{\sqrt{x}}\cdot\frac{1}{2}x^{-1/2}\). #Step 2: Find the second derivative# To find the second derivative, we need to differentiate \(\frac{dy}{dx}\) with respect to x again. We will need to use the product rule and chain rule here. The product rule is given by: \(\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\), where \(u = e^{\sqrt{x}}\), and \(v = \frac{1}{2}x^{-1/2}\).

Step 4: Find \(\frac{d}{dx}(e^{\sqrt{x}})\)

Applying the chain rule, we get: \(\frac{d}{dx}(e^{\sqrt{x}}) = e^{\sqrt{x}}\cdot\frac{1}{2}x^{-1/2}\).

Step 5: Find \(\frac{d}{dx}(x^{-1/2})\)

Taking the derivative of \(x^{-1/2}\) with respect to x, we get: \(\frac{d}{dx}(x^{-1/2}) = -\frac{1}{4}x^{-3/2}\). Now we can find \(\frac{d^2y}{dx^2}\):

Step 6: Find \(\frac{d^2y}{dx^2}\)

Using the product rule, we have: \(\frac{d^2y}{dx^2} = \frac{d}{dx}(e^{\sqrt{x}}\cdot \frac{1}{2}x^{-1/2})\) \(= \left(e^{\sqrt{x}}\right)\left(-\frac{1}{4}x^{-3/2}\right) + \left(\frac{1}{2}x^{-1/2}\right)\left(e^{\sqrt{x}}\cdot \frac{1}{2}x^{-1/2}\right)\) \(= -\frac{1}{4}e^{\sqrt{x}}x^{-3/2} + \frac{1}{4}e^{\sqrt{x}}x^{-1}\), which is the second derivative of the function \(y = e^{\sqrt{x}}\) with respect to x.

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of a composition of functions. Basically, when you have a function within another function, such as \( y = e^{\sqrt{x}} \), the chain rule is how we calculate the rate of change.
To apply the chain rule, you differentiate the outer function first, while keeping the inner function unchanged. In our exercise, the outer function is the exponential function \( e^u \) while the inner function is \( u = \sqrt{x} \).

• The outer derivative is \( \frac{dy}{du} = e^u \).
• Then, find \( \frac{du}{dx} = \frac{1}{2}x^{-1/2} \) since the inner function is \( \sqrt{x} = x^{1/2} \).

The product of these derivatives, \( \frac{dy}{du} \cdot \frac{du}{dx} \), gives \( \frac{dy}{dx} \). This step is crucial for finding higher-order derivatives later on.

Applying the Product Rule

The product rule is necessary when you're differentiating a product of functions. In our exercise, once you find \( \frac{dy}{dx} \), you need it again to reach \( \frac{d^2y}{dx^2} \).
The product rule formula is: \( \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \), where \( u \) and \( v \) are functions of \( x \).

• For \( y = e^{\sqrt{x}} \cdot \frac{1}{2}x^{-1/2} \), set \( u = e^{\sqrt{x}} \) and \( v = \frac{1}{2}x^{-1/2} \).
• The derivative \( \frac{du}{dx} \) uses the chain rule applied in the earlier step.
• The derivative \( \frac{dv}{dx} = -\frac{1}{4}x^{-3/2} \), essential for applying the product rule.

This systematic approach ensures you accurately capture the interaction between these multiplying functions.

Differentiating Exponential Functions

Differentiating exponential functions, especially with variable exponents, is common in calculus. In \( y = e^{\sqrt{x}} \), the base is \( e \), the exponential constant. Unlike constants, the derivative of such functions involves replicating the function itself.
Exponential functions have a unique property: the derivative \( \frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx} \). If \( u = \sqrt{x} \), as in our exercise, substitute it to find \( \frac{d}{dx}(e^{\sqrt{x}}) \). This results in multiplying the function by the derivative of the exponent (here, \( \frac{1}{2}x^{-1/2} \)).

• This multiplicative property simplifies differentiation, ensuring efficient work with exponential functions, even when combining them with other functions.
• For higher derivatives like \( \frac{d^2y}{dx^2} \), these repeated steps are key in combining techniques like chain and product rules effectively.

Embrace the power of this foundational differentiation rule to tackle complex expressions with confidence.