Question

Differentiate the following functions. $$ \begin{aligned} u &=\frac{1}{(a+b \cos x)^{2}} \\ \frac{d u}{d x} &=\sin x \\ \frac{d u}{d x} &=-\cos x \end{aligned} $$

Short answer

Question: Differentiate the following functions with respect to x: a) \(u=\frac{1}{(a+b\cos x)^2}\) b) \(u = \sin x\) c) \(u = -\cos x\) Answer: a) \(\frac{d u}{d x} = \frac{2b\sin x}{(a+b\cos x)^3}\) b) \(\frac{d u}{d x} = \cos x\) c) \(\frac{d u}{d x} = \sin x\)

Step-by-step solution

Differentiate the first function, \(u=\frac{1}{(a+b\cos x)^2}\)

To differentiate the first function with respect to x, we will apply the quotient rule, since the function can be expressed as the quotient of two functions. The quotient rule for differentiating a function \(\frac{f(x)}{g(x)}\) is given by the formula \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\). Here, \(f(x) = 1\) and \(g(x) = (a+b\cos x)^2\). We can rewrite the function as \(u(x) = 1 \div (a + b\cos x)^2\) and differentiate with respect to x. First, we need to find the derivatives of \(f(x)\) and \(g(x)\). The derivative of a constant w.r.t x is 0, therefore \(f'(x) = 0\). Now let's differentiate \(g(x) = (a+b\cos x)^2\). We can apply the chain rule, which gives us \(g'(x)=2(a+b\cos x)^1 \cdot (-b\sin x)\). Now we are ready to use the quotient rule formula: $$\frac{d u}{d x}=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$ Substituting the expressions, we get: $$\frac{d u}{d x}=\frac{0 \cdot (a+b\cos x)^2-1 \cdot 2(a+b\cos x)(-b\sin x)}{(a+b\cos x)^4}$$ Which simplifies to: $$\frac{d u}{d x}=\frac{2b\sin x}{(a+b\cos x)^3}$$

Differentiate the second function, \(u=\sin x\)

The derivative of \(\sin x\) with respect to x is a well-known result. The result is: $$\frac{d u}{d x}=\cos x$$

Differentiate the third function, \(u=-\cos x\)

The derivative of \(-\cos x\) with respect to x is also a well-known result. The result is: $$\frac{d u}{d x}=\sin x$$ In conclusion, the derivatives of the given functions are: $$ \begin{aligned} \frac{d u}{d x} &= \frac{2b\sin x}{(a+b\cos x)^3} \\ \frac{d u}{d x} &= \cos x \\ \frac{d u}{d x} &= \sin x \end{aligned} $$

Key concepts

Quotient Rule

When we need to differentiate a function that is a division, or a quotient, of two distinct functions, the Quotient Rule is our go-to tool. The rule is necessary because it allows us to differentiate such functions according to a structured method.

The Quotient Rule is expressed as:

• \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\)

In this formula, \(f(x)\) represents the numerator and \(g(x)\) the denominator. It requires the separate derivatives of both the numerator and the denominator.

In the example provided, our numerator function, \(f(x)\), was simply a constant 1, and the denominator, \(g(x) = (a+b\cos x)^2\). Calculating the derivatives and substituting them into the formula enabled us to find the derivative of the original function.

Remember, the quotient rule can manage any functions being divided. But it's important to carefully differentiate both components before applying the formula itself.

Chain Rule

The Chain Rule is a critical concept in calculus used for differentiating composite functions. A composite function involves an outer function and an inner function, such as \((a+b\cos x)^2\). Here, the outer function is the square, and the inner function is \(a+b\cos x\).

The Chain Rule is expressed as:

• \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

This rule helps in finding the derivative of complex functions by breaking them down into simpler parts. You differentiate the outer function first, taking the inner function as it is, then multiply by the derivative of the inner function.

In our example, we applied the Chain Rule to differentiate \((a+b\cos x)^2\). We first regarded it as \(2(a+b\cos x)^1\), then multiplied by the derivative of the inner function \(-b\sin x\). The result was used in the Quotient Rule to complete the derivative process efficiently.

• This method ensures accurate differentiation without omitting any components of the function.

Trigonometric Derivatives

Trigonometric functions like \(\sin x\) and \(\cos x\) have established derivatives that are commonly used in calculus.

Here are the primary derivatives for these functions:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \(-\sin x\).

In our exercise, we differentiated \(\sin x\) and \(-\cos x\) using these standard derivatives.

• For \(u = \sin x\), the derivative, \(\frac{du}{dx}\), is straightforward, yielding \(\cos x\).
• When differentiating \(-\cos x\), you multiply the derivative of \(\cos x\) by \(-1\), resulting in \(\sin x\).

These derivatives are fundamental tools in calculus, forming the basis for solving more complex trigonometric functions. Memorizing these derivatives is crucial as they recur frequently across various calculus problems.

Familiarity and practice will make these derivatives second nature, simplifying the process of differentiation involving trigonometric functions.