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Chapter 3: Problem 49

Derivatives of tower functions (or \(g^{h}\) ) Find the derivative of each function and evaluate the derivative at the given value of \(a\). $$f(x)=x^{\cos x} ; a=\pi / 2$$

Short Answer

Expert verified
Answer: The derivative of the function \(f(x) = x^{\cos x}\) at \(x = \frac{\pi}{2}\) is \(-\ln \frac{\pi}{2}\).

Step by step solution

01

Rewrite the function using logarithmic properties

We will apply the natural logarithm to both sides of the given function \(f(x) = x^{\cos x}\): $$\ln( f(x) ) = \ln( x^{\cos x} )$$ Using the logarithmic property \(\ln( x^y ) = y \cdot \ln( x )\), we can rewrite the right side of the equation as follows: $$\ln( f(x) ) = \cos x \cdot \ln x$$
02

Differentiate both sides of the equation with respect to x

Now, we can differentiate both sides of the equation with respect to \(x\) using the chain and product rules: $$\frac{1}{f(x)}\cdot\frac{d}{dx}f(x)= -\sin x \cdot \ln x + \cos x \cdot \frac{1}{x}$$ Now, by multiplying both sides of the equation by \(f(x)\), we can get the derivative of the function \(f(x)\): $$\frac{d}{dx}f(x)= f(x) \cdot (-\sin x \cdot \ln x + \cos x \cdot \frac{1}{x})$$ Since we are given the function \(f(x) = x^{\cos x}\), we can substitute this back into the equation above: $$\frac{d}{dx}f(x) = x^{\cos x} \cdot (-\sin x \cdot \ln x + \cos x \cdot \frac{1}{x})$$
03

Evaluate the derivative at the given value of x

Now that we have the derivative of the function, we can evaluate it at \(x = \frac{\pi}{2}\): $$\frac{d}{dx}f(x) \Big|_{x=\frac{\pi}{2}} = \left(\frac{\pi}{2}\right)^{\cos \frac{\pi}{2}} \cdot \left(-\sin \frac{\pi }{2} \cdot \ln \frac{\pi}{2} + \cos \frac{\pi}{2} \cdot \frac{1}{\frac{\pi}{2}}\right)$$ Since \(\cos{\frac{\pi }{2}}=0\) and \(\sin{\frac{\pi}{2}}=1\), the expression simplifies to: $$\frac{d}{dx}f(x)\Big|_{x=\frac{\pi}{2}} = 1^0 \cdot\left(-1 \cdot \ln \frac{\pi}{2} + 0 \cdot 2\right) = - \ln \frac{\pi}{2}$$ Therefore, the derivative of the function \(f(x) = x^{\cos x}\) at \(x = \frac{\pi}{2}\) is \(-\ln \frac{\pi}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Differentiation
Logarithmic differentiation is a powerful technique for finding the derivative of functions with exponents that are themselves functions, like tower functions. When faced with a function of the form \( g(x)^{h(x)} \), applying the logarithm to both sides simplifies the expression by bringing down the exponent as a multiplicative factor.

This approach turns a complex differentiation problem into a more manageable one by using properties of logarithms, such as \( \ln(x^y) = y \cdot \ln(x) \). After applying this property, the resulting equation can be differentiated using the product rule and the chain rule, making the process much easier than attempting to differentiate the function directly.
Chain Rule
The chain rule is a fundamental method in calculus used to find the derivative of composite functions. When a function consists of one function inside another, the chain rule is the tool you need.

Mathematically, if you have a function \( f(g(x)) \), the chain rule states that the derivative \( f'(g(x)) \) is equal to \( f'(g(x)) \cdot g'(x) \). This technique is used when differentiating the inner function and then multiplying it by the derivative of the outer function, allowing for the calculation of more complex derivatives like those involving tower functions.
Product Rule
The product rule comes into play when you need to find the derivative of a function that is the product of two other functions. For example, if you have a function \( u(x) \cdot v(x) \), the derivative is not simply the product of the derivatives of \( u \) and \( v \), but rather \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

This rule allows us to neatly find the derivative of each function individually and then combine them to get the total derivative. It's essential when dealing with logarithmic differentiation, as often the resulting equation is a product of functions that require the application of the product rule to successfully differentiate.
Evaluating Derivatives at a Point
After finding the derivative of a function, evaluating it at a specific point tells us the slope of the tangent line to the curve at that point. This is crucial for understanding the behavior of the function at a given value of \( x \).

To evaluate the derivative at a point, you simply substitute the value of \( x \) into the derivative equation. Often, this can involve simplifying expressions with trigonometric functions, especially when dealing with trigonometric exponents as seen with tower functions. The simplification process can greatly influence the resulting value and is an important step in fully solving differentiation problems.

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Most popular questions from this chapter

The volume of a torus (doughnut or bagel) with an inner radius of \(a\) and an outer radius of \(b\) is \(V=\pi^{2}(b+a)(b-a)^{2} / 4.\) a. Find \(d b / d a\) for a torus with a volume of \(64 \pi^{2}.\) b. Evaluate this derivative when \(a=6\) and \(b=10.\)

Carry out the following steps. \(x.\) a. Use implicit differentiation to find \(\frac{d y}{d x}.\) b. Find the slope of the curve at the given point. $$x y+x^{3 / 2} y^{-1 / 2}=2 ;(1,1)$$

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in kilometers, the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}\) a. Compute the pressure at the summit of Mt. Everest, which has an elevation of roughly \(10 \mathrm{km} .\) Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\) d. Does \(p^{\prime}(z)\) increase or decrease with \(z ?\) Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

The number of hours of daylight at any point on Earth fluctuates throughout the year. In the Northern Hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At \(40^{\circ}\) north latitude, the length of a day is approximated by $$D(t)=12-3 \cos \left(\frac{2 \pi(t+10)}{365}\right)$$ where \(D\) is measured in hours and \(0 \leq t \leq 365\) is measured in days, with \(t=0\) corresponding to January 1 a. Approximately how much daylight is there on March 1 \((t=59) ?\) b. Find the rate at which the daylight function changes. c. Find the rate at which the daylight function changes on March \(1 .\) Convert your answer to units of min/day and explain what this result means. d. Graph the function \(y=D^{\prime}(t)\) using a graphing utility. e. At what times of the year is the length of day changing most rapidly? Least rapidly?

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$e^{x y}=2 y$$
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