Question

Differentiate the following functions. $$ u=\frac{1}{\sin x+\cos x} $$

Short answer

Given the function \(u(x)=\frac{1}{\sin x+\cos x}\), compute the derivative with respect to x. First, rewrite the given function as \(u(x)=\frac{1}{h(x)}\) where \(h(x) = \sin x + \cos x\). Identify f(x) = 1 and g(x) = h(x) = \(\sin x + \cos x\). Differentiate both f(x) and g(x) with respect to x to get f'(x) = 0 and g'(x) = \(\cos x - \sin x\). Apply the quotient rule: \(u'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\) Substitute the expressions for f(x), f'(x), g(x), and g'(x) into the quotient rule: \(u'(x) = \frac{-\cos x +\sin x}{(\sin x + \cos x)^2}\) The completed solution is: \(u'(x) =\frac{-\cos x +\sin x}{(\sin x + \cos x)^2}\)

Step-by-step solution

Identify the functions f(x) and g(x)

In this case, we have \(u(x)=\frac{1}{\sin x+\cos x}\), which can be rewritten as \(\frac{1}{h(x)}\) where \(h(x) = \sin x + \cos x\). So we have \(f(x) = 1\) and \(g(x) = h(x) = \sin x + \cos x\).

Differentiate f(x) and g(x)

We need to find the derivatives of f(x) and g(x) with respect to x. As f(x) = 1, which is a constant, its derivative is: \(f'(x) = 0\). To find the derivative of g(x), we differentiate each term separately: $$g'(x) = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x)$$ As the derivative of \(\sin x\) is \(\cos x\) and the derivative of \(\cos x\) is \(-\sin x\), we get: $$g'(x) = \cos x - \sin x$$

Apply the Quotient Rule

Now we apply the quotient rule to find the derivative of u(x): $$u'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$ Substitute f(x), f'(x), g(x), and g'(x): $$u'(x) = \frac{0\cdot(\sin x + \cos x)-1\cdot(\cos x -\sin x)}{(\sin x + \cos x)^2}$$ As 0 times anything is 0, the equation simplifies to: $$u'(x) = \frac{-\cos x +\sin x}{(\sin x + \cos x)^2}$$ Our final answer for the derivative of the given function is: $$ u'(x) =\frac{-\cos x +\sin x}{(\sin x + \cos x)^2} $$

Key concepts

Quotient Rule

The quotient rule is a foundational concept in calculus for finding the derivative of a function that is presented as a quotient, or a ratio, of two other functions. Put simply, if you have a function defined as a division of two subfunctions, such as \( u(x) = \frac{f(x)}{g(x)} \), the quotient rule allows you to find the derivative of \( u(x) \) using the derivatives of \( f(x) \) and \( g(x) \) as follows:

\[ u'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]
By applying this rule, you differentiate the numerator and the denominator separately, and then combine them according to the formula. The quotient rule is particularly useful when you’re faced with complex rational expressions, and it is crucial to remember that the derivative of the bottom function is subtracted from the derivative of the top function.

Derivative of Sine Function

Understanding the derivative of the sine function is essential when working with trigonometric derivatives. The sine function, \( \text{sin}(x) \), is one of the basic trigonometric functions. In calculus, the derivative of \( \text{sin}(x) \) with respect to \( x \), often notated as \( \frac{d}{dx}\text{sin}(x) \), is simply the cosine function, \( \text{cos}(x) \).

\[ \frac{d}{dx}\text{sin}(x) = \text{cos}(x) \]
This relationship holds in classic calculus and is repeatedly used to solve calculus problems involving trigonometric functions. When a student is able to remember this straightforward derivative, it simplifies the process of finding the derivatives of more complex functions involving the sine function.

Derivative of Cosine Function

Similarly to the sine function, the cosine function, \( \text{cos}(x) \), is another trigonometric function frequently encountered in calculus problems. The derivative of \( \text{cos}(x) \) is a bit different, as it introduces the concept of the negative sine function. Mathematically, the derivative of the cosine function with respect to \( x \) is:

\[ \frac{d}{dx}\text{cos}(x) = -\text{sin}(x) \]
Meaning, when you differentiate \( \text{cos}(x) \), the result is the negative of the sine function. This highlights an important aspect of trigonometric functions, where each function is intricately related to the others through their derivatives. Memorizing the derivatives of these basic trigonometric functions is crucial for successfully tackling calculus problems that involve these elements.

Calculus Problems and Solutions

In calculus, solving problems often involves applying a variety of rules and formulas to find derivatives, integrals, and limits. Derivatives of functions, especially those involving trigonometric elements, are a common task in calculus. By understanding the rules, such as the quotient rule, product rule, and chain rule, along with the derivatives of basic functions, students can solve calculus problems effectively.

Working through exercises step by step, like in the given example of finding the derivative of a function involving the quotient of trigonometric functions, helps to build the knowledge and confidence required for more complex calculus solutions. Always start by identifying the functions involved, differentiate them separately, and apply the appropriate rule to find the desired derivative. Practice, combined with a strong grasp of fundamental concepts, is key to mastering calculus problems and solutions.