Question

Find the derivative. \(\frac{d}{d x} \int_{1}^{x^{3}} \sqrt{4-t^{2}} d t\)

Short answer

The derivative is \(3x^2 \sqrt{4-x^6}\).

Step-by-step solution

Recognize the Derivative of an Integral

To solve the derivative of an integral, observe that this is a problem dealing with the Fundamental Theorem of Calculus. Specifically, this problem involves the Fundamental Theorem of Calculus Part 2, which states that if you have an integral of the form \(\int_{a}^{g(x)} f(t)\, dt\), its derivative with respect to \(x\) is given by \(f(g(x)) \cdot g'(x)\).

Identify the Function Inside the Integral

For the integral \(\int_{1}^{x^{3}} \sqrt{4-t^{2}}\, dt\), identify the function \(f(t) = \sqrt{4-t^{2}}\) and the upper limit of integration \(g(x) = x^3\).

Apply the Fundamental Theorem of Calculus

Apply the derivative using the theorem. According to the theorem, the derivative of \(\int_{1}^{x^{3}} \sqrt{4-t^{2}}\, dt\) is \(\sqrt{4-(x^{3})^2} \cdot \frac{d}{dx}(x^3)\).

Compute the Derivative of the Upper Limit

Compute \(\frac{d}{dx}(x^3)\). This is \(3x^2\) since the derivative of \(x^n\) is \(nx^{n-1}\).

Simplify the Expression

Multiply the function evaluated at the upper limit by the derivative of the upper limit: \(\sqrt{4-x^6} \cdot 3x^2\). Thus, the derivative simplifies to \(3x^2 \sqrt{4-x^6}\).

Key concepts

derivative

In calculus, the concept of a derivative is essentially a tool to understand how a function changes as its input changes. The derivative provides the rate at which that function is changing at any given point. To put it simply, it tells us how fast something is happening. The notation can be written as \(\frac{d}{dx}(f(x))\), where we are finding the derivative of the function \(f\) with respect to \(x\). This concept is crucial as it helps in solving problems involving motion, optimization, and other situations where change is involved.

• The derivative is the slope of the tangent line to the function at a particular point.
• It tells us whether the function is increasing or decreasing at that point.
• Common formulas, like \(\frac{d}{dx}(x^n) = nx^{n-1}\), can be used to find derivatives easily.

Understanding derivatives is essential because they form the foundation of many calculus problems, like finding rates of change, and are used in fields ranging from physics to economics.

integral calculus

Integral calculus focuses on the accumulation of quantities, such as areas under curves, total distances, and more. It is essentially the reverse operation of taking a derivative. In particular, the definite integral \(\int_{a}^{b} f(t)\, dt\) represents the area under the curve \(f(t)\) from \(t = a\) to \(t = b\). The process of finding an integral is known as integration.

• Definite integrals yield a number that represents the area under the curve.
• Indefinite integrals involve finding a function whose derivative is the original function.
• The Fundamental Theorem of Calculus connects differentiation and integration, showing that integration can be reversed by differentiation.

Integral calculus helps us measure how quantities accumulate over a certain interval and plays a critical role in solving various real-world problems related to total size or change.

chain rule

The chain rule is a powerful technique in calculus used for finding the derivative of a composite function. In simpler terms, a composite function is where one function is inside another, such as \(f(g(x))\). To differentiate these, we apply the chain rule: take the derivative of the outer function and multiply it by the derivative of the inner function.

• The chain rule formula is \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\).
• This rule is particularly useful when dealing with nested functions.
• In solving complex problems, breaking them down using the chain rule can significantly simplify the calculation.

The chain rule is indispensable because it provides a straightforward method to differentiate even the most complex composite functions, which is crucial in calculus and its applications.