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Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{\cos x}\right)$$

Short Answer

Expert verified
Answer: The derivative of the function \(x^{\cos x}\) with respect to x is \(\frac{d}{dx}\left(x^{\cos x}\right) = x^{\cos x}\left((- \sin x) \cdot \ln(x) + \cos x \cdot \frac{1}{x}\right)\).

Step by step solution

01

Take the natural logarithm of both sides

Take the natural logarithm of the given function: $$\ln(y) = \ln\left(x^{\cos x}\right)$$ where \(y = x^{\cos x}\).
02

Simplify the logarithmic expression

Using the power rule of logarithms, we can simplify the right-hand side of the equation: $$\ln(y) = \cos x\cdot \ln(x)$$
03

Implicit differentiation with respect to x

Differentiate both sides of the equation with respect to x using the chain rule: $$\frac{1}{y}\cdot \frac{dy}{dx} = \frac{d}{dx}(\cos x\cdot\ln(x))$$
04

Differentiate the right-hand side

Apply the product rule to differentiate the right-hand side of the equation: $$\frac{1}{y}\cdot\frac{dy}{dx} = (-\sin x)\cdot \ln(x) + \cos x \cdot \frac{1}{x}$$
05

Solve for the desired derivative

Multiply both sides of the equation by y: $$\frac{dy}{dx} = y\left((- \sin x) \cdot \ln(x) + \cos x \cdot \frac{1}{x}\right)$$ Now replace y with \(x^{\cos x}\): $$\frac{dy}{dx} = x^{\cos x}\left((- \sin x) \cdot \ln(x) + \cos x \cdot \frac{1}{x}\right)$$ So the derivative of the given function is: $$\frac{d}{d x}\left(x^{\cos x}\right) = x^{\cos x}\left((- \sin x) \cdot \ln(x) + \cos x \cdot \frac{1}{x}\right)$$

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Key Concepts

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

Derivative Calculations
Derivative calculations are essential in calculus. They help us measure how a function changes at any point. In simple terms, a derivative gives the slope of the function's tangent line at a specific point. To calculate derivatives, we use various rules and techniques. Each technique depends on the function's form.

For the function \(x^{\cos x}\), a direct derivative approach is tricky. Why? Because the exponent is not constant but a function, \(\cos x\). So, we employ logarithmic differentiation. This method simplifies the process by using logarithms, making it easier to find derivatives of complicated expressions. Always keep in mind, when performing these calculations, the ultimate goal is to simplify the function's behavior at every point.
Implicit Differentiation
Implicit differentiation is a method used when it's hard to solve equations in terms of their dependent variables explicitly. Instead of rearranging everything, we take derivatives on both sides as is. It allows us to differentiate equations without needing to isolate \(y\). This strategy is super useful for complex equations.

For example, compare our case where \(y = x^{\cos x}\). By taking the natural logarithm of both sides, we worked with \(\ln(y) = \cos x \cdot \ln(x)\). We then used implicit differentiation to take the derivative of \(\ln(y)\) with respect to \(x\). Remember, always apply the chain rule. It allows for differentiating combined or nested functions systematically.
  • If you encounter a situation with \(y\), differentiate normally, then multiply by \(\frac{dy}{dx}\) for that component.
  • For terms not involving \(y\), apply regular differentiation rules as needed.
Implicit differentiation reduces complex derivatives to basic rules while keeping equations intact.
Power Rule of Logarithms
The power rule of logarithms is a valuable tool in differentiating complex expressions. It states that the logarithm of a power can be simplified. Namely, \(\ln(a^b) = b \cdot \ln(a)\). This rule is handy when you have an exponent that is a function itself.

In our exercise, the power rule transforms \(\ln(x^{\cos x})\) into \(\cos x \cdot \ln(x)\). This conversion makes differentiation easier by breaking down the problem into simpler multiplication and chain rules. Effectively, the power rule helps transform tricky expressions into manageable chunks. As we convert powers into products, it directly aligns with basic differentiation, which relies on straightforward operations. So, always remember, when faced with exponentials that are functions, let the power rule of logarithms simplify your task.

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Most popular questions from this chapter

Visualizing tangent and normal lines a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises \(63-68 .)\) b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right);\) \(\left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli)

Once Kate's kite reaches a height of \(50 \mathrm{ft}\) (above her hands), it rises no higher but drifts due east in a wind blowing \(5 \mathrm{ft} / \mathrm{s} .\) How fast is the string running through Kate's hands at the moment that she has released \(120 \mathrm{ft}\) of string?

A particle travels clockwise on a circular path of diameter \(R,\) monitored by a sensor on the circle at point \(P ;\) the other endpoint of the diameter on which the sensor lies is \(Q\) (see figure). Let \(\theta\) be the angle between the diameter \(P Q\) and the line from the sensor to the particle. Let \(c\) be the length of the chord from the particle's position to \(Q\) a. Calculate \(d \theta / d c\) b. Evaluate \(\left.\frac{d \theta}{d c}\right|_{c=0}\)

The Witch of Agnesi The graph of \(y=\frac{a^{3}}{x^{2}+a^{2}},\) where \(a\) is a constant, is called the witch of Agnesi (named after the 18th-century Italian mathematician Maria Agnesi). a. Let \(a=3\) and find an equation of the line tangent to \(y=\frac{27}{x^{2}+9}\) at \(x=2\) b. Plot the function and the tangent line found in part (a).

Visualizing tangent and normal lines a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises \(63-68 .)\) b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right);\) \(\left(x_{0}, y_{0}\right)=(2,2)\) (limaçon of Pascal)

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