Simplify The Following Expression D/dx Integral

Unlocking Calculus: How to Simplify d/dx ∫ (The Derivative of an Integral)

At the heart of calculus lies a beautiful and powerful connection between two of its central operations: differentiation and integration. The expression d/dx ∫ represents this link in its most concise form, and understanding how to simplify it is a fundamental skill for any student of mathematics, physics, or engineering. This seemingly cryptic notation is governed by the Fundamental Theorem of Calculus (FTC), a cornerstone that transforms the daunting task of evaluating integrals into a manageable process. Mastering this simplification rule not only streamlines problem-solving but also reveals the profound unity within calculus itself. This article will demystify the process, providing a clear, step-by-step guide to confidently handle expressions where a derivative and an integral are combined.

The Core Principle: The Fundamental Theorem of Calculus (Part 1)

The key to simplifying d/dx ∫ is the first part of the Fundamental Theorem of Calculus. It states that if you have a function f(t) that is continuous on an interval [a, b], and you define a new function F(x) as the definite integral of f from a constant a to a variable x:

F(x) = ∫ₐˣ f(t) dt

Then, the derivative of F(x) with respect to x is simply the original function f evaluated at the upper limit x.

d/dx [ ∫ₐˣ f(t) dt ] = f(x)

In Leibniz notation, this elegant result is written as:

d/dx ∫ₐˣ f(t) dt = f(x)

Why does this work? Intuitively, the integral ∫ₐˣ f(t) dt represents the accumulated area under the curve f(t) from t = a to t = x. When you take the derivative with respect to x, you are asking: "How does this accumulated area change as I nudge the upper limit x by a tiny amount?" The change in area is approximately the height of the function at x (f(x)) times the tiny width (dx). Dividing by dx to find the rate of change leaves you with f(x).

Key Takeaway:

When the upper limit of integration is the variable of differentiation (x) and the lower limit is a constant, the derivative of the integral is the integrand evaluated at that upper limit. The variable of integration inside the integral (traditionally t) is a "dummy variable" and disappears in the final answer.

Handling More Complex Scenarios

Real-world problems rarely present the simplest form. We must adapt the core rule for different situations.

1. Variable Lower Limit

What if the constant is on top and the variable is on the bottom? d/dx ∫ₓᵇ f(t) dt = ? Think of flipping the limits. Recall that ∫ₓᵇ f(t) dt = -∫ᵇₓ f(t) dt. Applying the FTC: d/dx [ -∫ᵇₓ f(t) dt ] = - [ f(x) ] = -f(x) Rule: A variable lower limit introduces a negative sign. The derivative is -f(x).

2. Both Limits Are Functions of x

This is the most common and important extension. Suppose we have: d/dx ∫_{u(x)}^{v(x)} f(t) dt Here, both the upper limit v(x) and the lower limit u(x) are functions of x. We use the Chain Rule in conjunction with the FTC.

1. Imagine temporarily holding the lower limit constant. The rate of change contributed by the moving upper limit v(x) is f(v(x)) * v'(x).
2. Now, imagine temporarily holding the upper limit constant. The rate of change contributed by the moving lower limit u(x) is -f(u(x)) * u'(x). (The negative sign appears for the same reason as in the variable lower limit case).
3. Combine these two effects.

General Formula (Leibniz Rule): d/dx ∫_{u(x)}^{v(x)} f(t) dt = f(v(x)) · v'(x) - f(u(x)) · u'(x)

This formula is your primary tool for simplifying complex expressions.

Worked Examples: From Simple to Advanced

Example 1 (Direct Application): Simplify: d/dx ∫₀ˣ cos(t²) dt

• Upper limit is x (variable), lower limit is 0 (constant).
• Apply FTC directly: f(x) = cos(x²).
• Answer: cos(x²)

Example 2 (Variable Lower Limit): Simplify: d/dx ∫ₓ⁵ e^{t³} dt

• This is d/dx [ -∫₅ₓ e^{t³} dt ].
• Apply rule: -f(x) = -e^{x³}.
• Answer: -e^{x³}

Example 3 (Both Limits Functions - The Chain Rule): Simplify: d/dx ∫_{sin(x)}^{x²} √(1 + t) dt

• Identify: u(x) = sin(x), v(x) = x², f(t) = √(1+t).
• Compute derivatives: u'(x) = cos(x), v'(x) = 2x.
• Apply Leibniz Rule: = f(v(x)) * v'(x) - f(u(x)) * u'(x) = √(1 + x²) * (2x) - √(1 + sin(x)) * cos(x)
• Answer: 2x√(1 + x²) - cos(x)√(1 + sin(x))

Example 4 (Nested Composition): Simplify: `d/dx ∫₁^{x