How To Find Dy/dx Of An Integral

Finding dy/dx of an Integral: A Step‑by‑Step Guide to Differentiation Under the Integral Sign

When a function is defined as an integral with respect to a variable, you often need to know how the function changes as that variable changes. Still, in calculus, this means computing the derivative dy/dx of an integral, which is a direct application of the Fundamental Theorem of Calculus and the Leibniz integral rule. Mastering this technique opens the door to solving complex problems in physics, engineering, and advanced mathematics That's the part that actually makes a difference..

Introduction

The expression

[ y = \int_{a}^{x} f(t),dt ]

defines y as the accumulated area under the curve f(t) from a fixed lower limit a to a variable upper limit x. The question “what is dy/dx of this integral?” is answered by the Fundamental Theorem of Calculus, Part 1, which states that the derivative of an integral with respect to its upper limit equals the integrand evaluated at that limit:

[ \frac{dy}{dx} = f(x). ]

On the flip side, many real‑world problems involve integrals where both limits are functions of x or where the integrand also depends on x. In those cases, the simple FTC result is insufficient, and we must use the Leibniz rule for differentiation under the integral sign. This article walks you through the theory, step‑by‑step procedures, and practical examples so you can confidently compute dy/dx for any integral you encounter Turns out it matters..

The Fundamental Theorem of Calculus (FTC)

The FTC provides the foundation for differentiating integrals:

1. If ( y = \int_{a}^{x} f(t),dt ) and f is continuous on ([a, x]), then

   [ \frac{dy}{dx} = f(x). ]

2. If the lower limit is also a function, say ( y = \int_{g(x)}^{h(x)} f(t),dt ), the FTC must be combined with the chain rule Still holds up..

The FTC is powerful because it tells us that integration and differentiation are inverse operations. When the limits are simple constants, the derivative is just the integrand evaluated at the variable limit.

The Leibniz Integral Rule (Differentiation Under the Integral Sign)

When the integral’s limits or integrand depend on x, we apply the Leibniz rule:

[ \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t),dt = f\bigl(x,\beta(x)\bigr),\beta'(x) ;-; f\bigl(x,\alpha(x)\bigr),\alpha'(x) ;+; \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}(x,t),dt . ]

Here:

• (\alpha(x)) and (\beta(x)) are the lower and upper limits.
• (f(x,t)) is the integrand, possibly depending on both x and t.
• (\beta'(x)) and (\alpha'(x)) are the derivatives of the limits.
• (\partial f/\partial x) denotes the partial derivative of the integrand with respect to x.

The rule has three intuitive parts:

1. Upper‑limit contribution – evaluate the integrand at the upper limit and multiply by its derivative.
2. Lower‑limit contribution – subtract the integrand at the lower limit times its derivative.
3. Integrand‑change contribution – integrate the partial derivative of the integrand over the interval.

Step‑by‑Step Procedure

1. Identify the integral form
   Write the integral as (\displaystyle I(x)=\int_{\alpha(x)}^{\beta(x)} f(x,t),dt).

2. Check the integrand
   Determine whether f depends on x (i.e., contains x outside the differential). If it does, you will need the partial‑derivative term Nothing fancy..

3. Differentiate the limits
   Compute (\alpha'(x)) and (\beta'(x)). If a limit is constant, its derivative is zero.

4. Apply the Leibniz formula
   Plug the pieces into the rule:

   [ I'(x)= f\bigl(x,\beta(x)\bigr)\beta'(x) - f\bigl(x,\alpha(x)\bigr)\alpha'(x) + \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}(x,t),dt . ]

5. Simplify
   Combine like terms, evaluate any definite integrals that remain, and write the final expression for dy/dx.

Example 1: Variable Upper Limit Only

Find (\displaystyle \frac{dy}{dx}) for

[ y = \int_{0}^{x} t^{2},dt . ]

• The integrand (f(t)=t^{2}) does not depend on x.
• Upper limit (\beta(x)=x) → (\beta'(x)=1).
• Lower limit (\alpha(x)=0) → (\alpha'(x)=0).

Using the Leibniz rule (or directly FTC):

[ \frac{dy}{dx}= f(x,\beta(x))\beta'(x) - f(x,\alpha(x))\alpha'(x) = x^{2}\cdot1 - 0 = x^{2}. ]

Thus, (\displaystyle \frac{dy}{dx}=x^{2}).

Example 2: Both Limits Are Functions

Compute (\displaystyle \frac{dy}{dx}) for

[ y = \int_{x^{2}}^{x^{3}} \sin(t),dt . ]

• Integrand (f(t)=\sin(t)) (no x dependence).
• (\beta(x)=x^{3}) → (\beta'(x)=3x^{2}).
• (\alpha(x)=x^{2}) → (\alpha'(x)=2x).

Apply the rule:

[ \frac{dy}{dx}= \sin(x^{3})\cdot 3x^{2} - \sin(x^{2})\cdot 2x . ]

So

[ \boxed{\displaystyle \frac{dy}{dx}=3x^{2}\sin(x^{3})-2x\sin(x^{2}) }. ]

Example 3: Integrand Depends on x (Full Leibniz Rule)

Find (\displaystyle \frac{dy}{dx}) for

[ y = \int_{1}^{x} e^{xt},dt . ]

Here the integrand contains x as a factor: (f(x,t)=e^{xt}) Simple as that..

• Upper limit (\beta(x)=x) → (\beta'(x)=1).
• Lower limit (\alpha(x)=1) → (\alpha'(x)=0).

Compute each term:

1. Upper‑limit term: (f(x,\beta(x))\beta'(x)=e^{x\cdot x}\cdot1 = e^{x^{2}}).

2. Lower‑limit term: zero (since (\alpha'(x)=0)).

3. Partial‑derivative term:

   [ \frac{\partial f}{\partial x}= \frac{\partial}{\partial x}e^{xt}=t,e^{xt}. ]

   The integral of this term is

   [ \int_{1}^{x} t,e^{xt},dt . ]

   This integral can be evaluated using integration by parts or recognized as the derivative of (\frac{1}{x}e^{xt}) with respect to t. Carrying out the integration yields

   [ \int_{1}^{x} t,e^{xt},dt = \frac{1}{x^{2}}\bigl(e^{x^{2}}-