Question

Find both first partial derivatives. \(f(x, y)=\int_{x}^{y}\left(t^{2}-1\right) d t\)

Short answer

The first partial derivatives of the given function are \(\frac{\partial f}{\partial x} = -(x^{2}-1)\) and \(\frac{\partial f}{\partial y} = (y^{2}-1)\).

Step-by-step solution

First Partial Derivative With Respect To \(x\)

Using the Fundamental Theorem of Calculus, the first partial derivative of \(f(x, y)\) with respect to \(x\) is obtained via direct substitution. Thus, \(\frac{\partial f}{\partial x} = -(x^{2}-1)\), where the negative sign results from deriving the lower limit.

First Partial Derivative With Respect To \(y\)

Next, to get the first partial derivative of \(f(x, y)\) with respect to \(y\), the same process applies, only this time, the derivative of the upper limit is taken. Hence, \(\frac{\partial f}{\partial y} = (y^{2}-1)\), with no negative sign, unlike the partial derivative with respect to \(x\).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical link between the concept of integration and differentiation, two major operations in calculus. In the context of partial derivatives, it allows us to compute derivatives of integrals with variable limits. This theorem can be applied in two parts:

• Part 1: It states that if function 'f' is continuous over an interval [a, b] and 'F' is the indefinite integral of 'f', then the definite integral of 'f' from a to b equals 'F(b) - F(a)'.
• Part 2: This part establishes that if 'f' is continuous over an interval and 'F' is any antiderivative of 'f' on that interval, then the derivative of 'F' with respect to 'x' is 'f(x)'.

These principles enable us to differentiate functions defined by an integral, as shown in the exercise provided. By applying the second part of the theorem, we see that the derivative of an integral with respect to its upper limit is the integrand evaluated at that limit, and the derivative with respect to the lower limit is the negative of the integrand evaluated at that limit.
This insight reveals why there's a change in sign when taking partial derivatives in our original exercise. The derivative with respect to the upper limit yielded a positive term, while the derivative with respect to the lower limit resulted in the negative term.

Partial Derivative With Respect to x

When taking the partial derivative of a function with respect to 'x', we treat 'y' as a constant. In the provided exercise, \(f(x, y)=\int_{x}^{y}\left(t^{2}-1\right) d t\), the Fundamental Theorem of Calculus guides us to differentiate the integral with respect to its lower limit 'x'.
When performing this operation, it's crucial to understand the impact of the limit of integration. Because 'x' is the lower limit of the integral, taking the partial derivative with respect to 'x' effectively applies the chain rule, resulting in the negative of the integrand, \(-\left(x^{2}-1\right)\). The negative is present due to the lower limit: as 'x' increases, the area under the curve, from 'x' to 'y', decreases, hence the minus sign in the derivative. This conceptual understanding is imperative for comprehending why the sign changes when differentiating with respect to the lower limit versus the upper limit.

Partial Derivative With Respect to y

On the other hand, taking the partial derivative with respect to 'y' treats 'x' as a constant and scrutinizes how the function changes when varying only 'y'. Referring back to our exercise, we look at the upper limit of the integral in \(f(x, y)=\int_{x}^{y}\left(t^{2}-1\right) d t\).
In this case, the partial derivative of the function with respect to 'y' is simply the integrand evaluated at 'y', which gives us \(\frac{\partial f}{\partial y} = (y^{2}-1)\).Unlike the partial derivative with respect to 'x', there is no negative sign since an increase in 'y' leads to an increase in the area under the curve from 'x' to 'y'. This encapsulates the essence of how varying 'y', the upper limit, affects the value of the integral - it directly reflects the value of the function at 'y', adjusting the overall area accordingly.