Question

Derivatives of integrals Simplify the following expressions. $$\frac{d}{d x} \int_{0}^{\cos x}\left(t^{4}+6\right) d t$$

Short answer

Question: Find the derivative of the integral function with respect to x: $$\frac{d}{d x} \int_{0}^{\cos x}\left(t^{4}+6\right) d t$$ Answer: The derivative of the integral function is: $$- \cos^4 x \sin x - 6 \sin x$$

Step-by-step solution

Identify the terms in the integral expression

Let's break down the given expression: $$\frac{d}{d x} \int_{0}^{\cos x}\left(t^{4}+6\right) d t$$ Here, a(x) = 0, b(x) = cos(x), and f(t) = t^4 + 6.

Find the derivatives of a(x) and b(x)

We need to find the derivatives of a(x) and b(x) with respect to x: $$\frac{d a(x)}{d x} = \frac{d(0)}{d x} = 0$$ $$\frac{d b(x)}{d x} = \frac{d(\cos x)}{d x} = -\sin x$$

Apply the Fundamental Theorem of Calculus Part 1

Now, we will apply the Fundamental Theorem of Calculus Part 1 to simplify the expression: $$\frac{d}{d x} \int_{0}^{\cos x}\left(t^{4}+6\right) d t = f(\cos x)\cdot\frac{d \cos x}{d x} - f(0)\cdot\frac{d(0)}{d x}$$

Substitute f(t), a(x), and b(x) derivatives

Substituting f(t) = t^4 + 6, and the derivatives of a(x) and b(x) we found in step 2: $$f(\cos x)\cdot\frac{d \cos x}{d x} - f(0)\cdot\frac{d(0)}{d x} = (\cos^4 x + 6)\cdot(-\sin x) - (0^4 + 6)\cdot 0$$

Simplify the expression

Simplify the expression: $$- \cos^4 x \sin x - 6 \sin x$$ Thus, the simplified expression for the derivative of the integral is: $$\frac{d}{d x} \int_{0}^{\cos x}\left(t^{4}+6\right) d t = - \cos^4 x \sin x - 6 \sin x$$

Key concepts

Derivative of Integral

In calculus, finding the derivative of an integral might initially seem tricky, but the process is made simpler using the Fundamental Theorem of Calculus. This theorem links the concept of differentiation with integration. When you're asked to find the derivative of an integral with variable limits, you're typically dealing with what is known as a "derivative of an integral" problem.

The Fundamental Theorem of Calculus states that if you have a function \( F(x) \) that is an integral with a variable upper limit, the derivative \( F'(x) \) with respect to \( x \) is simply the original function evaluated at the upper limit times the derivative of that upper limit.

For example, if you have an integral expression \( \int_{a(x)}^{b(x)} f(t) \, dt \), the derivative is:

• \( \frac{d}{dx} \int_{a}^{b(x)} f(t) \, dt = f(b(x)) \cdot b'(x) \)
• This result comes from applying the Fundamental Theorem Part 1 along with the chain rule when necessary.

Chain Rule in Calculus

The chain rule plays a vital role in differentiating integrals with variable bounds. Its job is crucial when taking the derivative of a composition of functions.

In our exercise, the integrand \( f(t) = t^4 + 6 \) isn't directly dependent on \( x \), but the upper limit \( b(x) = \cos(x) \) is, making the application of the chain rule necessary.

The chain rule states that if you have a function \( h(x) = g(f(x)) \), then its derivative \( h'(x) \) is \( g'(f(x)) \cdot f'(x) \). This helps you deal strategically with the derivative of a function nested within another function, which applies in our integral where the limit itself is a function of \( x \).

Thus, when you find \( \frac{d}{dx} \int_{0}^{\cos x} (t^4 + 6) \, dt \), you multiply the result of \( f(\cos x) \) by the derivative of \( \cos(x) \), taking the chain rule into account.

Calculus Problems

Calculus problems involving derivatives and integrals often require understanding several core concepts simultaneously. For students, it's crucial to build a robust grasp of both differentiation and integration principles, as well as techniques like substitution and parts.

In handling calculus problems like the one in this exercise, it boils down to:

• Identifying components within the integral like the function \( f(t) \) and variable limits \( a(x) \) and \( b(x) \).
• Determining when to apply the Fundamental Theorem of Calculus and the chain rule effectively.
• Substituting and simplifying to arrive at the final derivative form, \( -\cos^4 x \sin x - 6 \sin x \) in this case.

Problem-solving in calculus involves practice, patience, and the ability to link the theories to various types of equations, enhancing a student's skill to tackle a broad range of calculus questions efficiently.