Question

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

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

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: $$\ln y = \ln\left(x^{\left(x^{10}\right)}\right)$$

Use Logarithm Properties

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

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: $$\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left(x^{10}\cdot\ln x\right)$$

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: $$\frac{1}{y}\frac{dy}{dx} = 10x^9\cdot\ln x + x^{10}\cdot\frac{1}{x}$$

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

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

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: $$\boxed{\frac{d}{dx}\left(x^{\left(x^{10}\right)}\right)=x^{\left(x^{10}\right)}\left(10x^9\cdot\ln x + x^{9}\right)}$$

Key concepts

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 \(x\).