Problem 32 - Single Variable Calculus

Question

Find the derivative of the function. $$ y=\cos \left(x^{2}-3 x+1\right)+\tan \left(\frac{2}{x}\right) $$

Step-by-Step Solution

Answer

The derivative of the given function is: \[ \frac{dy}{dx} = -\sin \left(x^{2}-3 x+1\right) \cdot (2x - 3) - \frac{2 \sec^2 \left(\frac{2}{x}\right)}{x^2}. \]

Step 1: Differentiate the first term

We should start by differentiating the first term, which is the cosine function. Apply the chain rule to the function \[ y_{1} = \cos \left(x^{2}-3 x+1\right). \] To find \(\frac{dy_{1}}{dx}\), we first find the derivative of the outer function (the cosine) with respect to the inner function (the polynomial), and then multiply it by the derivative of the inner function with respect to x. So, we obtain: \[ \frac{dy_{1}}{dx} = -\sin \left(x^{2}-3 x+1\right) \cdot (2x - 3). \]

Step 2: Differentiate the second term

Now, let's differentiate the second term, which is the tangent function. Apply the chain rule and the quotient rule to the function \[ y_{2} = \tan \left(\frac{2}{x}\right). \] To find \(\frac{dy_{2}}{dx}\), we first find the derivative of the outer function (the tangent) with respect to the inner function (the fraction), and then multiply it by the derivative of the inner function with respect to x. We obtain: \[ \frac{dy_{2}}{dx} = \sec^2 \left(\frac{2}{x}\right) \cdot \frac{d}{dx}\left(\frac{2}{x}\right). \] Now we need to find \(\frac{d}{dx}\left(\frac{2}{x}\right)\) by applying the quotient rule: \[ \frac{d}{dx}\left(\frac{2}{x}\right) = \frac{-2}{x^2}. \] Now, multiply the two parts together: \[ \frac{dy_{2}}{dx} = \sec^2 \left(\frac{2}{x}\right) \cdot \frac{-2}{x^2}. \]

Step 3: Combine the derivatives of the two terms

Now we need to combine the derivatives of the two terms to find the derivative of the whole function: \[ \frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx} = -\sin \left(x^{2}-3 x+1\right) \cdot (2x - 3) + \sec^2 \left(\frac{2}{x}\right) \cdot \frac{-2}{x^2}. \] So, the derivative of given function is: \[ \frac{dy}{dx} = -\sin \left(x^{2}-3 x+1\right) \cdot (2x - 3) - \frac{2 \sec^2 \left(\frac{2}{x}\right)}{x^2}. \]

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of a function that is composed of other functions. When you have a function within another function, the chain rule helps us unravel that complexity. In mathematical terms, if you have a function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).
For example, in the case of the function \( y_1 = \cos(x^2 - 3x + 1) \), the chain rule is applied by differentiating the outer function \( \cos \) with respect to the implicit inner function \( x^2 - 3x + 1 \), resulting in

• \( \frac{d}{dx}(\cos(u)) = -\sin(u) \cdot \frac{du}{dx} \)
• where \( u = x^2 - 3x + 1 \).

This process simplifies complex derivatives, tackling them step-by-step.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are foundational in calculus, especially when finding derivatives. The derivatives of these functions often appear in real-world applications, such as physics and engineering. Here are the basic derivatives you need to remember:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

In our exercise, these identities come into play. For instance, when differentiating \( \cos(x^2 - 3x + 1) \) and \( \tan(\frac{2}{x}) \), we use these basic derivative rules as part of applying the chain rule.

Product Rule

The product rule is not explicitly used in this specific exercise, but it's important to understand when differentiating products of two functions. If you have two functions multiplied together, \( u(x) \cdot v(x) \), the derivative is calculated using:\[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]This rule ensures every part of the product is accounted for when differentiating. The concept is more visible in functions composed of multiple products, explaining the individual contributions (changes) of each component.

Quotient Rule

The quotient rule is vital when differentiating fractions of two functions, such as \( \frac{2}{x} \) in our example. It provides a method to find the derivative of the division of two functions without simplifying into products. If you have a function \( \frac{u(x)}{v(x)} \), the derivative is:\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}\]In the exercise, differentiating \( \frac{2}{x} \) employs this rule, where

• \( u(x) = 2 \), \( v(x) = x \).
• \( u'(x) = 0 \), \( v'(x) = 1 \).

Applying the quotient rule gives you the derivative in a structured way, especially when manually working through complex expressions.