Question

Berechnen Sie die Ableitung und den Gradienten der Funktion $$ f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=x_{1} \ln \left(x_{2} x_{3}\right)+e^{x_{2}+x_{1} x_{3}} $$

Short answer

The gradient is \( \nabla f = \left( \ln(x_2 x_3) + x_3 e^{x_2 + x_1 x_3}, \frac{x_1}{x_2} + e^{x_2 + x_1 x_3}, \frac{x_1}{x_3} + x_1 e^{x_2 + x_1 x_3}, 0 \right) \).

Step-by-step solution

Understand the Function

First, understand the function given as \( f(x_1, x_2, x_3, x_4) = x_1 \ln(x_2 x_3) + e^{x_2 + x_1 x_3} \). This is a multi-variable function with variables \( x_1, x_2, x_3, \) and \( x_4 \). The function contains a logarithmic term and an exponential term.

Differentiate with Respect to \(x_1\)

Find the partial derivative of \(f\) with respect to \(x_1\). Use the chain rule for the terms involved:\[ \frac{\partial f}{\partial x_1} = \ln(x_2 x_3) + x_3 e^{x_2 + x_1 x_3} \].

Differentiate with Respect to \(x_2\)

Find the partial derivative of \(f\) with respect to \(x_2\). Use the chain and product rule as necessary:\[ \frac{\partial f}{\partial x_2} = \frac{x_1}{x_2 x_3} \cdot x_3 + e^{x_2 + x_1 x_3} \]. Simplifying gives \( \frac{\partial f}{\partial x_2} = \frac{x_1}{x_2} + e^{x_2 + x_1 x_3} \).

Differentiate with Respect to \(x_3\)

Find the partial derivative of \(f\) with respect to \(x_3\). Again, apply the product and chain rule:\[ \frac{\partial f}{\partial x_3} = \frac{x_1 x_2}{x_2 x_3} + x_1 e^{x_2 + x_1 x_3} \]. Simplifying gives \( \frac{\partial f}{\partial x_3} = \frac{x_1}{x_3} + x_1 e^{x_2 + x_1 x_3} \).

Differentiate with Respect to \(x_4\)

As \(x_4\) does not appear in the function, the partial derivative with respect to \(x_4\) is zero:\[ \frac{\partial f}{\partial x_4} = 0 \].

Formulate the Gradient Vector

Combine these partial derivatives into the gradient vector of \(f\): \[ abla f = \left( \ln(x_2 x_3) + x_3 e^{x_2 + x_1 x_3}, \frac{x_1}{x_2} + e^{x_2 + x_1 x_3}, \frac{x_1}{x_3} + x_1 e^{x_2 + x_1 x_3}, 0 \right) \].

Key concepts

Gradient Vector

The gradient vector is an essential concept in multivariable calculus. It provides the direction and rate of the fastest increase of a function. For a function of several variables, the gradient is a vector composed of all the partial derivatives of that function. Each component in the gradient corresponds to the partial derivative with respect to one variable.

• In our exercise, the function \( f(x_1, x_2, x_3, x_4) \) is differentiated to find the partial derivatives with respect to each variable.
• The gradient vector \( abla f \) is then formed from these derivatives.
• This results in \[ abla f = \left( \ln(x_2 x_3) + x_3 e^{x_2 + x_1 x_3}, \frac{x_1}{x_2} + e^{x_2 + x_1 x_3}, \frac{x_1}{x_3} + x_1 e^{x_2 + x_1 x_3}, 0 \right) \]

The gradient tells us how the function changes in different directions.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of more than one variable. Instead of dealing with functions of a single variable \( x \), we analyze functions of variables like \( x_1, x_2, x_3, \) and \( x_4 \) simultaneously.

• This allows us to study how changes in multiple variables impact the value of a function.
• It leads to concepts such as partial derivatives, which are crucial for understanding complex systems and changes.

In the exercise, multivariable calculus is used to navigate through partial derivatives with respect to each variable.

Chain Rule

The chain rule is a fundamental tool in calculus and plays a crucial role in differentiating composite functions. Simply put, it allows us to differentiate a function that is dependent on another function.

• When we have a composite function, the chain rule helps us find the derivative of each part.
• This is especially useful in multivariable functions where one variable may depend on others.

In our solution, the chain rule was applied to the exponential and logarithm terms of the function, facilitating the calculation of partial derivatives.

Product Rule

The product rule is another helpful differentiation rule. It is used when finding the derivative of a product of two functions. The rule states that if you have two differentiable functions \( u(x) \) and \( v(x) \), their derivative is given by:
\[ (uv)' = u'v + uv' \]

• In multivariable calculus, the product rule helps handle terms where variables multiply each other.
• During the calculation of partial derivatives of the given function, the product rule was employed effectively.

Utilizing the product rule enables us to handle the interactions between different variables within the function efficiently.