Problem 10 - Analysis 1: Ein Lehr- und Arbeitsbuch für Studienanfänger

Question

a) Begründen sie, daß die Funktion \(x \longmapsto \sin (\cos x), x \in \mathbb{R}\) differenzierbar ist und berechnen sie die Ableitung. b) Berechnen sie die Ableitungen der differenzierbaren Funktionen 1) \(x \mapsto\left(x^{2}+\cos x\right)^{2}, \quad x \in R\); 2) \(x \mapsto a^{x}, x \in \mathbb{R},(a>0, a \neq 1) ;\) 3) \(x \mapsto \sqrt{\frac{x^{2}+x+1}{3 x-9}}, x>3\)

Step-by-Step Solution

Answer

a) The derivative is \(-\sin x \cdot \cos(\cos x)\). b1) \(2(x^2 + \cos x)(2x - \sin x)\). b2) \(a^x \ln(a)\). b3) Use the quotient rule and chain rule for \(\sqrt{y}\).

Step 1: Checking Differentiability of the Function

To show that the function \(x \mapsto \sin(\cos x)\) is differentiable, we must first note that both \(\sin\) and \(\cos\) are differentiable functions on \(\mathbb{R}\). The composition of differentiable functions is also differentiable. Therefore, \(\sin(\cos x)\) is differentiable on \(\mathbb{R}\).

Step 2: Finding the Derivative of the Composition

To find \(\frac{d}{dx}[\sin(\cos x)]\), we apply the chain rule. Let \(u = \cos x\), then \(\sin(u)\) is \(\sin(\cos x)\). The derivative is \(\cos(u)\) times the derivative of \(u\), which is \(-\sin x\). Thus, the derivative is \(-\sin x \cdot \cos(\cos x)\).

Step 3: Finding the Derivative of the First Function (a)

The function is \(x \mapsto (x^2 + \cos x)^2\). We use the chain rule. Let \(v = x^2 + \cos x\), so our function is \(v^2\). The derivative of \(v^2\) is \(2v \cdot v'\) where \(v' = 2x - \sin x\). The derivative is \(2(x^2 + \cos x)(2x - \sin x)\).

Step 4: Finding the Derivative of the Exponential Function (b)

The function is \(x \mapsto a^x\). The derivative of \(a^x\) is \(a^x \ln(a)\) since the exponential function has the derivative of itself times the natural log of its base. Thus, \(\frac{d}{dx}[a^x] = a^x \ln(a)\).

Step 5: Finding the Derivative of the Rational Function (c)

The function is \(x \mapsto \sqrt{\frac{x^2 + x + 1}{3x - 9}}\). Let \(y = \frac{x^2 + x + 1}{3x - 9}\), so \(\sqrt{y}\). The derivative \(\frac{d}{dx}[y^{1/2}] = \frac{1}{2\sqrt{y}}y'\). Use the quotient rule for \(y'\): \(\frac{(2x+1)(3x-9) - (x^2 + x + 1)(3)}{(3x-9)^2}\). Simplify to get the derivative.

Key Concepts

Chain Rule

The Chain Rule is a vital concept in calculus. It allows us to find the derivative of composite functions. To apply the Chain Rule, identify the "outer" and "inner" functions. Differentiate the outer function and multiply it by the derivative of the inner function.

For example, with the function \( \sin(\cos x) \), \( \sin \) is the outer function and \( \cos x \) is the inner. By applying the Chain Rule, differentiate \( \sin \) to get \( \cos \), and multiply it by the derivative of \( \cos x \), which is \(-\sin x\). Hence, we get \(-\sin x \cdot \cos(\cos x)\).

Keep in mind that this rule helps in breaking down complex differentiation tasks into manageable steps. Once you become comfortable identifying the layers of composed functions, the Chain Rule becomes an incredibly powerful tool in your mathematical toolkit.

Composite Functions

Understanding composite functions is crucial in calculus. A composite function is formed when one function is applied to the result of another function. They are usually represented as \( f(g(x)) \), where \( g(x) \) is inside \( f \).

In the exercise, the derivative needed involved a composite function of the form \( \sin(\cos x) \), where \( \cos x \) is inserted into \( \sin \). The core process involves working from the inside out using methods like the Chain Rule.

By studying these patterns, it becomes easier to tackle more intricate functions. Consider these steps as forming an assembly line for functions, where each piece must be processed in order!

Exponential Functions

Exponential functions are prevalent in mathematics and science. They can model growth and decay processes. In calculus, differentiating exponential functions involves applying unique rules.

For any function of the form \( a^x \), where \( a \) is a constant, the derivative is \( a^x \ln(a) \). This arises because the function grows by a constant factor, and the natural log captures this growth rate.

It’s significant to remember this special derivative rule, as it differentiates exponential functions from polynomials or other trigonometric functions. Always include the \( \ln(a) \) term as this distinguishes the rate of change specific to the base \( a \).

Quotient Rule

In calculus, when you need to find the derivative of a function that is the quotient of two differentiable functions, the Quotient Rule becomes crucial. It provides a way to systematically handle these derivatives.

The rule states: if you have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable, the derivative \( y' \) is given by:

• \( y' = \frac{v u' - u v'}{v^2} \)

This formula effectively subtracts the product of the derivative of the numerator and the denominator from the product of the numerator and the derivative of the denominator, all over the square of the denominator.

In our exercise, the function was a square root of a quotient, but breaking it down required first differentiating the inner fraction using the Quotient Rule. This helps resolve complex expressions step-by-step by following structured rules.