Question

Find the following derivatives: (i) \(\frac{d}{d x} \int_{a}^{b} \sin \left(x^{2}\right) d x\) (ii) \(\frac{d}{d a} \int_{a}^{x^{2}} \sin \left(x^{2}\right) d x\) (iii) \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{0}^{\mathrm{x}^{2}} \sqrt{1+\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}\) (iv) \(\frac{d}{d x} \int_{x^{2}}^{x^{3}} \frac{d t}{\sqrt{1+t^{2}}}\) (v) \(\frac{d}{d x} \int_{x^{2}}^{x^{3}} \frac{d x}{\sqrt{1+x^{2}}}\) (vi) \(\frac{\mathrm{d}}{\mathrm{dx}} \int_{\mathrm{t}^{2}}^{\mathrm{x}^{3}} \frac{\mathrm{dt}}{\sqrt{\mathrm{x}^{2}+\mathrm{t}^{4}}}\)

Short answer

Question: **Derivative of \(\frac{d}{dx} \int_{a}^{b} \sin(x^{2})dx\)** Solution: Since the limits of integration, \(a\) and \(b\), are constants, the derivative is zero: \(\frac{d}{dx} \int_a^b \sin(x^2) dx = 0\) Question: **Derivative of \(\frac{d}{da} \int_{a}^{x^{2}} \sin(x^{2}) dx\)** Solution: Using the Fundamental Theorem of Calculus and Chain Rule, the derivative is \(-\sin(a^2)\). Question: **Derivative of \(\frac{d}{dx} \int_{0}^{x^{2}} \sqrt{1+x^{2}} dx\)** Solution: The derivative is \(2x\sqrt{1+x^4}\). Question: **Derivative of \(\frac{d}{dx} \int_{x^{2}}^{x^{3}} \frac{dt}{\sqrt{1+t^{2}}}\)** Solution: The derivative is \(\frac{3x^2}{\sqrt{1+(x^3)^2}} - \frac{2x}{\sqrt{1+(x^2)^2}}\). Question (not well-defined): **Derivative of \(\frac{d}{dx} \int_{x^{2}}^{x^{3}} \frac{dx}{\sqrt{1+x^{2}}}\)** Solution: This question is not well-defined since the integral should be in terms of a different variable, such as 't'. Question: **Derivative of \(\frac{d}{dx} \int_{t^{2}}^{x^{3}} \frac{dt}{\sqrt{x^{2}+t^{4}}}\)** Solution: The derivative is \(\frac{3x^2}{\sqrt{x^2+(x^3)^4}}\).

Step-by-step solution

Apply the Fundamental Theorem of Calculus

The derivative will change the lower limit to be in terms of `t`, so we have \(\frac{d}{da} \int_a^{x^2} \sin(t^2) dt\)

Apply the Chain Rule

Now applying the chain rule, we have \(-\sin(a^2)\) #(iii) Derivative of \(\frac{d}{dx} \int_{0}^{x^{2}} \sqrt{1+x^{2}} dx\)

Apply the Fundamental Theorem of Calculus

The derivative will change the upper limit to be in terms of `t`, so we have \(\frac{d}{dx} \int_0^{x^2} \sqrt{1+t^2} dt\)

Apply Chain Rule

Now applying the chain rule, we have \(2x\sqrt{1+x^4}\) #(iv) Derivative of \(\frac{d}{dx} \int_{x^{2}}^{x^{3}} \frac{dt}{\sqrt{1+t^{2}}}\)

Apply Leibniz Rule

Using the Leibniz Rule, which is \(\frac{d}{dx} \int_{x^2}^{x^3} f(t)dt = f(x^3)\cdot(3x^2) - f(x^2)\cdot(2x)\), we have \(\frac{3x^2}{\sqrt{1+(x^3)^2}} - \frac{2x}{\sqrt{1+(x^2)^2}}\) #(v) Derivative of \(\frac{d}{dx} \int_{x^{2}}^{x^{3}} \frac{dx}{\sqrt{1+x^{2}}}\) The variable inside the integral and the derivative are the same, which is incorrect. This question is not well-defined, as the integral should be in terms of a different variable (for example, 't'). #(vi) Derivative of \(\frac{d}{dx} \int_{t^{2}}^{x^{3}} \frac{dt}{\sqrt{x^{2}+t^{4}}}\)

Apply Leibniz Rule

Using the Leibniz Rule, which is \(\frac{d}{dx} \int_{t^2}^{x^3} f(t)dt = f(x^3)\cdot(3x^2) - f(t^2)\cdot(0)\), we have \(\frac{3x^2}{\sqrt{x^2+(x^3)^4}}\) Note that in question (v), the notation was incorrect and the problem could not be solved.

Key concepts

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of integral calculus and serves as a bridge between differentiation and integration. It is split into two parts: the first part provides a link between the antiderivative and the integral, stating that if a function is continuous over an interval, then the integral of that function over that interval is equal to the difference between the values of its antiderivative at the endpoints of the interval. The second part, which is often the focus for derivative of integral problems, states that if you have a function that’s continuous on an interval and an antiderivative of that function, the derivative of the integral from a constant to a variable upper limit x equals the integrand (the function being integrated) evaluated at x.

For example, in the exercise \(\frac{d}{dx}\int_{0}^{x^{2}}\sqrt{1+x^{2}}dx\), applying the Fundamental Theorem of Calculus, one would evaluate the integrand \(\sqrt{1+t^{2}}\) at the upper limit \(x^{2}\), resulting in \(\sqrt{1+(x^{2})^{2}}\). This provides a method to find the derivative of the integral without directly integrating the function.

The Chain Rule in Calculus

The Chain Rule is a formula to compute the derivative of a composite function. In essence, if a variable u is a function of v, which in turn is a function of w, then the rate of change of u with respect to w can be found by multiplying the rate of change of u with respect to v by the rate of change of v with respect to w. Symbolically, this is expressed as \(\frac{du}{dw} = \frac{du}{dv} \cdot \frac{dv}{dw}\).

In the context of derivatives of integrals, the Chain Rule becomes particularly useful when the limits of the integral are functions of x, rather than constants. For the exercise \(\frac{d}{da}\int_{a}^{x^{2}}\sin(t^{2})dt\), after applying the Fundamental Theorem of Calculus, the Chain Rule is needed to handle the derivative of the lower limit of integration, which is a function of a. The resulting derivative is the product of the derivative of the integrand with respect to 'a' and the derivative of 'a' with respect to 'x'.

The Leibniz Rule

Leibniz Rule is a generalization of the Fundamental Theorem of Calculus for functions with variable limits. It describes how to differentiate an integral where one or both of the limits are functions rather than constants. The rule says that if the integral of f(t) from g(x) to h(x) is considered, then the derivative of this integral with respect to x is given by the integrand evaluated at the upper limit function, multiplied by the derivative of this upper limit function, minus the integrand evaluated at the lower limit function, multiplied by the derivative of this lower limit function.

Mathematically, it's represented as \(\frac{d}{dx}\int_{g(x)}^{h(x)}f(t)dt = f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x)\). In our exercise \(\frac{d}{dx}\int_{x^{2}}^{x^{3}}\frac{dt}{\sqrt{1+t^{2}}}\), applying Leibniz Rule simplifies the process of finding the derivative by directly substituting the limits into the integrand and differentiating them, thus avoiding the need to directly integrate the function. It handles the complexity of varying limits in the integral, offering a streamlined method for solving such problems.