Question

If \(x=\int_{1}^{t^{2}} z \ln z d z\) and \(y=\int_{t^{2}}^{1} z^{2} \ln z d z\) find \(\frac{d y}{d x}\).

Short answer

Question: Find the derivative of y with respect to x, where the functions x and y are given as definite integrals: $$x(t) = \int_{1}^{t^{2}} z \ln z d z$$ $$y(t) = \int_{t^{2}}^{1} z^{2} \ln z d z$$ Answer: The derivative of y with respect to x, denoted as dy/dx, equals to -1.

Step-by-step solution

Apply the fundamental theorem of calculus

By applying the fundamental theorem of calculus, we can find the functions x(t) and y(t): $$ x(t) = \int_{1}^{t^{2}} z \ln z d z \implies \frac{d x(t)}{dt} = t^{2} \ln (t^{2})\cdot 2t $$ $$ y(t) = \int_{t^{2}}^{1} z^{2} \ln z d z \implies \frac{d y(t)}{-dt} = t^{2} \ln (t^{2})\cdot 2t $$ Note that dy(t)/-dt because the limits are from t^2 to 1. If the limits were 1 to t^2, we would keep dy(t)/dt.

Apply the chain rule

Now, we find the derivative of x and y with respect to t, using the chain rule: $$ \frac{dx}{dt} = 2t^3 \ln (t^2) \implies \frac{dx}{dt} = 4t^3 \ln t $$ $$ \frac{dy}{-dt} = 2t^3 \ln (t^2) \implies \frac{dy}{dt} = -4t^3 \ln t $$

Calculate dy/dx

Now, to find the derivative of y with respect to x, we will divide dy/dt by dx/dt: $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-4t^3 \ln t}{4t^3 \ln t} = -1 $$ So, the value of dy/dx is -1.

Key concepts

Chain Rule in Calculus

The Chain Rule is a fundamental tool in calculus for finding the derivative of a composite function. It's particularly useful when one function is nested inside another. To apply the chain rule, imagine you have two functions, one inside the other. Let’s call them f(g(x)). When taking the derivative of the composite function with respect to x, you first differentiate the outer function f with respect to the inner function g, and then multiply by the derivative of the inner function g with respect to x. In formal terms, if h(x) = f(g(x)), then the derivative h'(x) is given by f'(g(x)) ∗ g'(x).

For example, if you need to differentiate h(x) = (2x^3 - x)^5, you would first differentiate the outer function, which in this case is raising something to the fifth power, and then multiply by the derivative of the inner function, which is 2x^3 - x. This would yield h'(x) = 5(2x^3 - x)^4 ∗ (6x^2 - 1).

The Chain Rule is especially pertinent when dealing with integrals because it allows for the substitution of variables, simplifying the integration process by transforming a complex integral into a easier one before integrating.

Integrals in Calculus

Integrals are another cornerstone of calculus and provide a way to calculate the area under curves, among other things. The process of integration is the reverse of taking a derivative, and it’s based on accumulating the infinitesimal contributions of a function's values over an interval. There are two main types of integrals: indefinite and definite integrals. The indefinite integral, or antiderivative, represents a family of functions and includes a constant of integration since differentiation wipes out any constant. On the other hand, a definite integral computes the net area between the function and the x-axis over an interval from a to b.

When integrating functions with respect to x, we often use the power rule, which in its basic form states that the integral of x^n is x^(n+1)/(n+1) plus a constant, assuming n is not equal to -1. More complex functions may require methods like substitution, integration by parts, or partial fractions. Furthermore, the Fundamental Theorem of Calculus directly links the process of differentiation with that of integration. This theorem tells us that if F is an antiderivative of a function f on an interval [a, b], then the definite integral of f from a to b is given by F(b) - F(a).

Derivative with Respect to x

When we talk about taking the derivative of a function with respect to x, we are looking for the rate at which the function changes as x changes. This is a crucial concept in calculus, as derivatives represent how a quantity changes in response to changes in another quantity. The notation for the derivative of y with respect to x is written as dy/dx or y'(x). To compute this, we use different rules of differentiation, such as the power rule, product rule, quotient rule, and as mentioned before, the chain rule.

For instance, if you have a simple function like y = 3x^2, the derivative dy/dx is found by bringing down the exponent as a coefficient (using the power rule) and then reducing the exponent by one. So, the derivative would be dy/dx = 2 ∗ 3x^(2-1) = 6x. If the function is more complex or involves multiple variables, partial derivatives may be involved, where one variable is differentiated while keeping others constant. Derivatives are not only foundational for calculus but also for numerous applications in science, engineering, economics, and beyond, wherever rates of change are critical.