Question

If \(z=\int_{\sin x}^{\cos x} \frac{\sin t}{t} d t\), find \(\frac{d z}{d x}\)

Short answer

\( \frac{d z}{d x} = - \frac{\sin (\cos x) \cdot \sin x}{\cos x} - \frac{\sin (\sin x) \cdot \cos x}{\sin x} \).

Step-by-step solution

Understand the integral representation

The given function is defined as an integral: \[ z = \int_{\sin x}^{\cos x} \frac{\sin t}{t} \ dt \].

Apply Leibniz Rule

To find the derivative \( \frac{d z}{d x} \), we need to use the Leibniz rule for differentiation under the integral sign. Leibniz rule states: \[ \frac{d}{d x} \left( \int_{a(x)}^{b(x)} f(t) \ dt \right) = f(b(x)) \frac{d b(x)}{d x} - f(a(x)) \frac{d a(x)}{d x} \].

Identify the upper and lower limits

In this problem, the upper limit is \( b(x) = \cos x \) and the lower limit is \( a(x) = \sin x \).

Differentiate the limits

Compute the derivatives of the limits with respect to \( x \): \[ \frac{d(b(x))}{d x} = \frac{d(\cos x)}{d x} = -\sin x \] \[ \frac{d(a(x))}{d x} = \frac{d(\sin x)}{d x} = \cos x \].

Substitute into Leibniz Rule

Substitute the limits and their derivatives into the Leibniz Rule. The integrand is \( \frac{\sin t}{t} \). Hence:\[ \frac{d z}{d x} = \frac{\sin (\cos x)}{\cos x} \cdot (-\sin x) - \frac{\sin (\sin x)}{\sin x} \cdot \cos x \].

Simplify the expression

Simplify the expression to obtain the final derivative:\[ \frac{d z}{d x} = - \frac{\sin (\cos x) \cdot \sin x}{\cos x} - \frac{\sin (\sin x) \cdot \cos x}{\sin x} \].

Key concepts

integral limits

Integral limits define the bounds within which the integration is performed. In our exercise, the limits are functions of x: the upper limit is given by \(\text{cos}\thinspace x\) and the lower limit by \(\text{sin}\thinspace x\). Understanding these limits is crucial in the process of differentiation under integral signs.
To summarize:

• Upper limit: \(b(x) = \text{cos}\thinspace x\)
• Lower limit: \(a(x) = \text{sin}\thinspace x\)

These bounds change with x, which significantly impacts how we find the derivative of the integral function.

differentiation under integral sign

Differentiation under the integral sign is a powerful technique allowing us to find the derivative of an integral whose limits are functions themselves. Leibniz rule provides a systematic way to accomplish this:
\[ \frac{d}{d x} \bigg( \thinspace \text{int}_{a(x)}^{b(x)} f(t) \thinspace dt \bigg) = f(b(x)) \frac{d b(x)}{d x} - f(a(x)) \frac{d a(x)}{d x} \thinspace. \]
This rule involves:

• Evaluating the integrand at the upper and lower limits.
• Multiplying these evaluations by the derivatives of the respective limits.

In practical terms, for our function \(z = \text{int}_{\text{sin}\thinspace x}^{\text{cos}\thinspace x} \frac{\text{sin} \thinspace t}{t} \thinspace dt\),
we substitute \(b(x) = \text{cos}\thinspace x\) and \(a(x) = \text{sin}\thinspace x\) to apply the Leibniz rule effectively.

trigonometric functions

Trigonometric functions are an essential branch of mathematics, often appearing in calculus for their properties and their role in defining periodic phenomena. In our exercise, functions like \( \text{sin}\thinspace x \) and \( \text{cos}\thinspace x \) play a crucial role in both the integral limits and the actual differentiation:

• For the lower limit \(a(x) = \text{sin}\thinspace x\), its derivative is \( \text{cos}\thinspace x \).
• For the upper limit \(b(x) = \text{cos}\thinspace x\), its derivative is \( -\text{sin}\thinspace x \).

Moreover, the integrand of our function also involves \( \text{sin}\thinspace t\), making it vital to understand how these trigonometric functions behave and interact. Their derivatives dictate how we differentiate more complex expressions where trigonometric functions set the boundaries or appear within the integrand.