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Partial derivatives Find the first partial derivatives of the following functions. $$f(x, y)=\int_{x}^{y^{3}} e^{t} d t$$

Short Answer

Expert verified
Answer: The first partial derivatives are: $$\frac{\partial f}{\partial x} = e^x$$ $$\frac{\partial f}{\partial y} = 3y^2 e^{y^3}$$

Step by step solution

01

Use the Fundamental Theorem of Calculus to take the derivative of the integral

We are given the function \(f(x, y)=\int_{x}^{y^{3}} e^{t} dt\). First, let's find the derivative of the integral, using the Fundamental Theorem of Calculus. For a function \(F(x) = \int_{a(x)}^{b(x)} g(t) dt\), we have: $$\frac{dF}{dx} = g(b(x)) \cdot \frac{db}{dx} - g(a(x)) \cdot \frac{da}{dx}$$ In our case, \(a(x) = x\) \(b(x) = y^3\) \(g(t) = e^t\) Now we can use the Fundamental Theorem of Calculus to differentiate \(f(x, y)\).
02

Differentiate f(x, y) with respect to x (∂f/∂x)

Using the formula from Step 1, we differentiate \(f(x, y)\) with respect to x: $$\frac{\partial}{\partial x} \int_{x}^{y^{3}} e^{t} dt = e^x \cdot \frac{\partial x}{\partial x} - e^{y^3} \cdot \frac{\partial y^3}{\partial x}$$ Since \(\frac{\partial x}{\partial x} = 1\) and \(\frac{\partial y^3}{\partial x} = 0\) (as x and y are independent variables), we have: $$\frac{\partial}{\partial x} \int_{x}^{y^{3}} e^{t} dt = e^x$$ So the first partial derivative with respect to x is: $$\frac{\partial f}{\partial x} = e^x$$
03

Differentiate f(x, y) with respect to y (∂f/∂y)

Now, we differentiate \(f(x, y)\) with respect to y: $$\frac{\partial}{\partial y} \int_{x}^{y^{3}} e^{t} dt = e^{y^3} \cdot \frac{\partial y^3}{\partial y} - e^x \cdot \frac{\partial x}{\partial y}$$ Since \(\frac{\partial x}{\partial y} = 0\) and \(\frac{\partial y^3}{\partial y} = 3y^2\), we have: $$\frac{\partial}{\partial y} \int_{x}^{y^{3}} e^{t} dt = e^{y^3} \cdot 3y^2$$ So the first partial derivative with respect to y is: $$\frac{\partial f}{\partial y} = 3y^2 e^{y^3}$$
04

Conclusion

The first partial derivatives of the given function are: $$\frac{\partial f}{\partial x} = e^x$$ $$\frac{\partial f}{\partial y} = 3y^2 e^{y^3}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus serves as a bridge between integration and differentiation, two principal concepts in calculus. It is divided into two parts; the first part provides an explicit connection between the derivative and the integral. The theorem states that if you have a continuous function, say \( g(t) \), over an interval \[a, b\], and you define another function \( F(x) \) where \( F(x) = \int_{a}^{x} g(t) \, dt \) for \( x \in \[a, b\], \) then \( F(x) \) is differentiable on \( (a, b) \) and \( F'(x) = g(x) \) for every \( x \in (a, b). \) This powerful result allows you to compute the derivative of an integral without directly evaluating the integral itself, which is particularly useful in problems involving partial derivatives of functions defined by integrals, as was seen in the given exercise.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions of several variables. In this broader context, differentiation includes partial derivatives, gradients, and directional derivatives. A partial derivative measures how a function changes as only one variable changes while the others are held constant. For instance, given a function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) focuses on the rate of change of \( f \) with respect to \( x \), when \( y \) is considered constant. Multivariable calculus plays a vital role in various fields such as physics, engineering, and economics, becoming crucial in modeling systems with several variables at play. The exercise provided beautifully illustrates the calculation of the first partial derivatives, a fundamental aspect of multivariable calculus.
Integration
Integration is the process of finding the integral of a function, which is the inverse operation of differentiation. It is used to compute areas, volumes, central points, and many useful things, but it's not just for solving 'area under a curve' problems. Integration can represent the accumulation of quantities, such as distance traveled over time, and is central to many aspects of mathematics and physics. In the context of finding the partial derivatives in our exercise, integration sets the stage for the function we're working with before we apply differentiation via the Fundamental Theorem of Calculus.
Exponential Functions
Exponential functions are powerful and appear across various disciplines from compounding bank interest to population growth modeling. The general form of an exponential function is \( f(x) = a^x \), where \( a \) is a positive constant. One particularly important exponential function for calculus, as well as for our exercise, is \( e^x \) where \( e \) is the base of the natural logarithm, approximately equal to 2.71828. This function has unique properties, such as being its own derivative and integral, which makes it extremely useful in solving differential equations and studying change models. In the given exercise, the exponential function \( e^t \) is integrated over an interval, leading to a function defined by an integral, where the properties of exponential functions are essential in finding the partial derivatives.

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