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General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{\left(x^{10}\right)}\right)$$

Short Answer

Expert verified
Question: Find the derivative of the function \(x^{\left(x^{10}\right)}\) using logarithmic differentiation. Answer: The derivative of the function \(x^{\left(x^{10}\right)}\) is \(\frac{d}{dx}\left(x^{\left(x^{10}\right)}\right)=x^{\left(x^{10}\right)}\left(10x^9\cdot\ln x + x^{9}\right)\).

Step by step solution

01

Apply Natural Logarithm to both sides

To apply logarithmic differentiation, let's use the natural logarithm (ln) and denote the function as \(y = x^{\left(x^{10}\right)}\). We take the natural logarithm of both sides: $$\ln y = \ln\left(x^{\left(x^{10}\right)}\right)$$
02

Use Logarithm Properties

Apply the logarithmic exponent property to simplify the expression: $$\ln y = x^{10} \cdot \ln x$$
03

Differentiate Both Sides

Now, differentiate both sides of the equation with respect to \(x\): $$\frac{d}{dx}(\ln y) = \frac{d}{dx}\left(x^{10}\cdot\ln x\right)$$ Remember, when differentiating \(\ln y\) with respect to \(x\), we need to use the chain rule, as \(y\) is a function of \(x\). We get: $$\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left(x^{10}\cdot\ln x\right)$$
04

Differentiate the Right Side of the Equation

To differentiate the right side, we use the product rule: $$(u\cdot v)' = u'\cdot v + u\cdot v'$$ Let \(u = x^{10}\) and \(v =\ln x\). Differentiate them with respect to \(x\): $$u' = \frac{d}{dx}(x^{10}) = 10x^9$$ $$v' = \frac{d}{dx}(\ln x) = \frac{1}{x}$$ Apply the product rule: $$\frac{1}{y}\frac{dy}{dx} = 10x^9\cdot\ln x + x^{10}\cdot\frac{1}{x}$$
05

Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y\) to find the derivative: $$\frac{dy}{dx} = y\left(10x^9\cdot\ln x + x^{10}\cdot\frac{1}{x}\right)$$
06

Replace y with the Original Function

Finally, substitute the original function back into the equation for \(y\): $$\frac{dy}{dx} = x^{\left(x^{10}\right)}\left(10x^9\cdot\ln x + x^{10}\cdot\frac{1}{x}\right)$$ And that's the derivative of the given function: $$\boxed{\frac{d}{dx}\left(x^{\left(x^{10}\right)}\right)=x^{\left(x^{10}\right)}\left(10x^9\cdot\ln x + x^{9}\right)}$$

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Key Concepts

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

Natural Logarithm
When you encounter a complicated expression like \(x^{(x^{10})}\), traditional differentiation techniques might not directly apply. This is where the natural logarithm, often represented as \(\ln\), becomes invaluable. The natural logarithm is the inverse of the exponential function. In mathematical terms, if \(y = e^x\), then \(x = ln(y)\). Its unique properties allow us to simplify complex equations.

A particularly useful property is the logarithmic identity \(\ln(a^b) = b\cdot\ln(a)\), which helps transform multiplicative relationships into additive ones, thus simplifying differentiation. In the exercise, applying \(\ln\) transforms the power function into a product, enabling the use of rules for differentiation that are easier to manage.
Product Rule
The product rule is essential when differentiating expressions where two functions are multiplied by each other, as seen in the form \(u(x)\cdot v(x)\). The rule states that the derivative of the product of two functions \(u\) and \(v\) is the derivative of the first function \(u\) times the second function \(v\) plus the first function \(u\) times the derivative of the second function \(v\). Expressed mathematically, the product rule is \((u\cdot v)' = u'\cdot v + u\cdot v'\).

In our exercise, after applying logarithmic differentiation, the product rule enables us to handle the derivative of \(x^{10}\cdot \ln(x)\), neatly separating the two parts to be differentiated individually and then combined for a solution.
Chain Rule
Another indispensable tool in calculus is the chain rule, which is used for differentiating composite functions. A composite function can be thought of as a function within another function—imagine putting one function inside the other. Mathematically, if \(f(x) = h(g(x))\), then the chain rule tells us that \(f'(x) = h'(g(x))\cdot g'(x)\).

In our specific problem, the chain rule comes into play when differentiating \(\ln(y)\) with respect to \(x\), where \(y\) itself is a function of \(x\). This is essential since \(y\) represents our complex original function \(x^{(x^{10})}\), and we need to differentiate it with respect to \(x\), not \(y\). The chain rule allows us to do so logically and systematically.
Exponential Derivatives
Differentiating exponential functions often involves the natural base \(e\), but can be extended to other bases using logarithmic differentiation. For standard exponential functions of the form \(a^x\) where \(a\) is a constant, the derivative is \((a^x)' = a^x\cdot\ln(a)\), where the natural logarithm of \(a\) appears as a multiplier.

However, when the exponent includes the independent variable \(x\), as in our exercise, this calls for a more complex approach. Logarithmic differentiation transforms the original function into a form where the techniques for derivatives of exponentials can be properly utilized. As a result, we obtain the derivative expressed in terms of the original function with an additional algebraic expression involving \(x\) and the natural logarithm of \(x\).

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Most popular questions from this chapter

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(x y=a ; x^{2}-y^{2}=b,\) where \(a\) and \(b\) are constants

Assuming that \(f\) is differentiable for all \(x,\) simplify \(\lim _{x \rightarrow 5} \frac{f\left(x^{2}\right)-f(25)}{x-5}.\)

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

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. 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)

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