Chapter 4: Problem 14
Bestimmen Sie die Ableitung der Funktion \(z=\tan (x y)\) l?ngs der Kurve \(y=x^{3}\).
Short Answer
Expert verified
The derivative is \( \frac{dz}{dx} = 4x^3 \text{sec}^2(x^4) \).
Step by step solution
01
- Express the function along the curve
Given the function is defined as \( z = \tan(xy) \) and the curve is \( y = x^3 \). Substitute \( y = x^3 \) into the function \( z \). Thus, \( z = \tan(x \times x^3) = \tan(x^4) \).
02
- Differentiate the function with respect to x
Differentiate \( z = \tan(x^4) \) with respect to \( x \). Use the chain rule: \( \frac{dz}{dx} = \frac{d}{dx}[\tan(x^4)] = \frac{d}{dx}[u] \times \frac{d}{dx}[x^4] \) where \( u = \tan(x^4) \).
03
- Apply the chain rule
Using the chain rule, note that \( \frac{d}{dx}[\tan(x^4)] = \frac{1}{\tan^2(x^4) + 1} \times \frac{d}{dx}[x^4] \). Since \( \frac{d}{dx}[x^4] = 4x^3 \), the expression becomes \( \frac{dz}{dx} = \frac{1}{\tan^2(x^4) + 1} \times 4x^3 \).
04
- Simplify the expression
Using the identity \( \tan^2(x) + 1 = \text{sec}^2(x) \), the expression simplifies to \( \frac{dz}{dx} = 4x^3 \times \text{sec}^2(x^4) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ableitung
The concept of 'Ableitung' in calculus refers to the derivative of a function. A derivative represents the rate at which a function is changing at any given point. It's essential in understanding how a function behaves incrementally. For instance, if you have a function like \( z = \tan(xy) \), the process to find its derivative requires several steps and techniques from calculus.
Understanding derivatives involves:
Understanding derivatives involves:
- Recognizing the function you need to derive.
- Using rules and formulae such as the product rule, quotient rule, and chain rule to find the derivative.
- Simplifying the resulting expressions.
Kettenregel
The 'Kettenregel' or chain rule is a fundamental rule in calculus used to differentiate a composite function. If a function, say \(z\), is given as \( z = \tan(x^4) \), applying the chain rule is crucial to finding its derivative.
The chain rule states that if you have two functions, \(u(x)\) and \(v(x)\), such that \(y = u(v(x))\), then the derivative of \(y\) with respect to \(x\) is:
\[ \frac{dy}{dx} = \frac{du}{dv} \times \frac{dv}{dx} \] In simpler terms, you first find the derivative of the outer function, then multiply it with the derivative of the inner function.
For the function \( z = \tan(x^4) \), applying the chain rule means differentiating \( \tan(u) \) where \( u = x^4 \), and then differentiating \( x^4 \). Consequently, we apply:
\[ \frac{d}{dx}[\tan(x^4)] = \frac{1}{\tan^2(x^4) + 1} \times \frac{d}{dx}[x^4] \] and,
\[ \frac{d}{dx}[x^4] = 4x^3 \]
The chain rule states that if you have two functions, \(u(x)\) and \(v(x)\), such that \(y = u(v(x))\), then the derivative of \(y\) with respect to \(x\) is:
\[ \frac{dy}{dx} = \frac{du}{dv} \times \frac{dv}{dx} \] In simpler terms, you first find the derivative of the outer function, then multiply it with the derivative of the inner function.
For the function \( z = \tan(x^4) \), applying the chain rule means differentiating \( \tan(u) \) where \( u = x^4 \), and then differentiating \( x^4 \). Consequently, we apply:
\[ \frac{d}{dx}[\tan(x^4)] = \frac{1}{\tan^2(x^4) + 1} \times \frac{d}{dx}[x^4] \] and,
\[ \frac{d}{dx}[x^4] = 4x^3 \]
Kurvendiskussion
'Kurvendiskussion', or curve sketching, is the process of analyzing and sketching the graph of a function. This usually involves finding:
- The function's domain and range
- Intercepts with the axes
- Critical points and inflection points
- Behavior at infinity and asymptotes
- Intervals where the function is increasing or decreasing
- Concavity and points of inflection
tan Funktion
The 'tan' function, short for 'tangent', is a trigonometric function that can be expressed as:
\[ \tan(x) = \frac{\text{sin}(x)}{\text{cos}(x)} \] The tangent function is periodic with a period of \( \theta = \frac{\text{Ï€}}{2} \) and exhibits undefined points, vertical asymptotes, at multiples of \( \frac{\text{Ï€}}{2} \).
When dealing with derivatives of the tangent function, the result is:
\[ \frac{d}{dx}[\tan(x)] = \text{sec}^2(x) \] Note that this relationship plays a pivotal role in our exercise where differentiating \( \tan(x^4) \) leads to:
\[ \frac{d}{dx}[\tan(u)] = \text{sec}^2(u) \times \frac{du}{dx} \]
\[ \tan(x) = \frac{\text{sin}(x)}{\text{cos}(x)} \] The tangent function is periodic with a period of \( \theta = \frac{\text{Ï€}}{2} \) and exhibits undefined points, vertical asymptotes, at multiples of \( \frac{\text{Ï€}}{2} \).
When dealing with derivatives of the tangent function, the result is:
\[ \frac{d}{dx}[\tan(x)] = \text{sec}^2(x) \] Note that this relationship plays a pivotal role in our exercise where differentiating \( \tan(x^4) \) leads to:
\[ \frac{d}{dx}[\tan(u)] = \text{sec}^2(u) \times \frac{du}{dx} \]
sec Funktion
The 'sec' function, or secant function, is another trigonometric function related to the cosine function:
\[ \text{sec}(x) = \frac{1}{\text{cos}(x)} \] The secant function's behavior is tied closely to the cosine function and is undefined wherever the cosine is zero.
In terms of differentiation, the derivative of the sec function is:
\[ \frac{d}{dx}[\text{sec}(x)] = \text{sec}(x) \times \text{tan}(x) \] This becomes particularly useful when simplifying expressions involving \( \tan(x) \) or \( \tan(x^4) \) as it shows up in our chain rule utilization where:
\[ \text{sec}^2(x^4) = \tan^2(x^4) + 1 \]
\[ \text{sec}(x) = \frac{1}{\text{cos}(x)} \] The secant function's behavior is tied closely to the cosine function and is undefined wherever the cosine is zero.
In terms of differentiation, the derivative of the sec function is:
\[ \frac{d}{dx}[\text{sec}(x)] = \text{sec}(x) \times \text{tan}(x) \] This becomes particularly useful when simplifying expressions involving \( \tan(x) \) or \( \tan(x^4) \) as it shows up in our chain rule utilization where:
\[ \text{sec}^2(x^4) = \tan^2(x^4) + 1 \]