Question

Find \(\frac{d y}{d x}\). $$y=e^{x} \sqrt{x^{3}}$$

Short answer

Question: Find the derivative of the function \(y = e^x \sqrt{x^3}\). Answer: The derivative of the given function is \(\frac{dy}{dx} = e^{x}x^{\frac{1}{2}}\left(\frac{3}{2} + x^{\frac{3}{2}}\right)\).

Step-by-step solution

Identify the functions

In this exercise, we have two functions: 1. The exponential function \(u(x) = e^{x}\) 2. The radical function \(v(x) = \sqrt{x^{3}}\)

Find derivatives of both functions

We need to find the derivatives of both, \(u(x)\) and \(v(x)\). 1. Derivative of exponential function \(u(x) = e^{x}\): $$\frac{du}{dx} = \frac{d(e^{x})}{dx} = e^{x}$$ 2. Derivative of radical function \(v(x) = \sqrt{x^{3}}\). We can rewrite it as \(x^{\frac{3}{2}}\), and then find the derivative: $$\frac{dv}{dx} = \frac{d(x^{\frac{3}{2}})}{dx} = \frac{3}{2}x^{\frac{1}{2}}$$

Apply product rule

Now we need to use the product rule to find the derivative of \(y = u(x)v(x)\). According to the product rule: $$\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$$

Calculate the derivative

Substituting the values of \(u(x)\), \(v(x)\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) from steps 1 and 2 into the product rule formula: $$\frac{dy}{dx} = e^x \left(\frac{3}{2}x^{\frac{1}{2}}\right) + \sqrt{x^3} (e^x)$$

Simplify the expression

Now we can combine the terms to simplify the expression: $$\frac{dy}{dx} = e^{x}x^{\frac{1}{2}}\left(\frac{3}{2} + x^{\frac{3}{2}}\right)$$ The derivative of the given function is: $$\frac{dy}{dx} = e^{x}x^{\frac{1}{2}}\left(\frac{3}{2} + x^{\frac{3}{2}}\right)$$

Key concepts

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In our function, the exponential part is represented by \( e^x \). This base \( e \) is an irrational number approximately equal to 2.718. It's called Euler's number, commonly used because it has unique properties, especially in calculus. The beauty of exponential functions like \( e^x \) is that their derivative is the same as the function itself. So, when we differentiate \( e^x \), the result is still \( e^x \). This simplicity is one reason exponential functions are prominent in many fields, including finance and the natural sciences. These functions typically model growth processes, such as population growth, radioactive decay, and compound interest, due to their continuous and proportional growth characteristics.

Product Rule

The product rule is a fundamental technique in calculus used to differentiate products of two functions. It enables us to tackle problems where two dependent variables need differentiation simultaneously. The rule is stated mathematically as follows: If you have two differentiable functions \( u(x) \) and \( v(x) \), then the derivative of their product \( y = u(x)v(x) \) is given by:

• \( \frac{d}{dx}[u(x) \cdot v(x)] = u(x) \cdot \frac{dv}{dx} + v(x) \cdot \frac{du}{dx} \).

In our scenario, we identified \( u(x) = e^x \) and \( v(x) = \sqrt{x^3} \). By employing the product rule, we are seamlessly able to find the derivative of these two intersecting functions. This rule is handy when solving problems involving the multiplication of different working components, making it a cornerstone concept for many calculus operations.

Radical Function

Radical functions are expressions involving roots, such as square roots or higher, of a variable. Our function \( v(x) = \sqrt{x^3} \) is a classic example, specifically a square root. It's often helpful to rewrite radicals using exponents to simplify differentiation and integration. For instance, \( \sqrt{x^3} \) can be rewritten as \( x^{3/2} \), which is more intuitive for calculus operations.

• The derivative of a radical function follows the power rule: if \( v(x) = x^n \), its derivative, \( \frac{dv}{dx} = nx^{n-1} \).

In this exercise, differentiating \( x^{3/2} \) gives \( \frac{3}{2}x^{1/2} \). Radical functions often appear in scenarios involving distance or area calculations, and converting them to power functions facilitates easier calculus operations.