Question

Derivative of \(u(x)^{v(x)}\) Use logarithmic differentiation to prove that $$\frac{d}{d x}\left(u(x)^{v(x)}\right)=u(x)^{v(x)}\left(\frac{d v}{d x} \ln u(x)+\frac{v(x)}{u(x)} \frac{d u}{d x}\right)$$.

Short answer

Question: Find the derivative of a function in the form \(u(x)^{v(x)}\). Answer: The derivative of the function \(u(x)^{v(x)}\) is given by: $$\frac{d}{d x}\left(u(x)^{v(x)}\right)=u(x)^{v(x)}\left(\frac{d v}{d x} \ln{[u(x)]}+\frac{v(x)}{u(x)} \frac{d u}{d x}\right)$$

Step-by-step solution

Write down the given function and apply natural logarithm to both sides

First, let's define the function we have to find the derivative of: $$y(x) = u(x)^{v(x)}$$ To use logarithmic differentiation, we will now take the natural logarithm of both sides of the equation. $$\ln{[y(x)]} = \ln{[u(x)^{v(x)}]}$$

Simplify the right side using logarithm properties

We can simplify the right side of the equation using the property of logarithms that states that \(\ln{a^b} = b \ln a\). This gives us: $$\ln{[y(x)]} = v(x)\ln{[u(x)]}$$

Implicitly differentiate both sides with respect to \(x\)

Now, we will differentiate both sides of the equation with respect to \(x\): $$\frac{d}{dx}\ln{[y(x)]}=\frac{d}{dx}\left[v(x)\ln{[u(x)]}\right]$$ Using the chain rule, we have: $$\frac{y'(x)}{y(x)}=\frac{d v}{d x}\ln{[u(x)]}+v(x)\frac{1}{u(x)}\frac{d u}{d x}$$

Multiply both sides by \(y(x)\) and use the original definition of the function

In this step, we will multiply both sides of the equation by \(y(x)\). This will give us the derivative we were looking for, \(y'(x)\). Recall that \(y(x) = u(x)^{v(x)}\): $$y'(x) = u(x)^{v(x)}\left(\frac{d v}{d x}\ln{[u(x)]}+v(x)\frac{1}{u(x)}\frac{d u}{d x}\right)$$

Write the final result

The derivative of the function \(u(x)^{v(x)}\) is given by: $$\frac{d}{d x}\left(u(x)^{v(x)}\right)=u(x)^{v(x)}\left(\frac{d v}{d x} \ln{[u(x)]}+\frac{v(x)}{u(x)} \frac{d u}{d x}\right)$$

Key concepts

Implicit Differentiation

Implicit differentiation is an essential technique in calculus when dealing with equations that define functions implicitly rather than explicitly. Instead of a straightforward function like \( y = f(x) \), we often encounter equations where \( y \) cannot be isolated, for instance, equations that define a relation between \( x \) and \( y \) without solving for one variable in terms of the other. When differentiating such relations, we treat \( y \) as an implicit function of \( x \), effectively allowing us to differentiate \( y \) with respect to \( x \).

This process involves using the chain rule to differentiate \( y \) terms, as if \( y \) were an explicit function of \( x \). It's particularly useful in logarithmic differentiation, as shown in the solution to the given exercise: we implicitly differentiate both sides of the equation after applying the natural logarithm to the function \( u(x)^{v(x)} \), treating \( y(x) \) as an implicit function.

Chain Rule

The chain rule is a fundamental rule in differentiation that allows us to compute the derivative of composite functions. When we have a function \( h(x) \) that can be decomposed into two functions such that \( h(x) = f(g(x)) \), the derivative of \( h \) with respect to \( x \) is the product of the derivative of \( f \) with respect to \( g \) and the derivative of \( g \) with respect to \( x \), or \( h'(x) = f'(g(x)) \cdot g'(x) \).

In the context of logarithmic differentiation, we often encounter situations where we need to differentiate an expression that is the product of two functions, which is exactly where the chain rule comes in handy. As seen in our example, when differentiating \( v(x){[u(x)]} \), we use the chain rule to differentiate \( v(x) \) and \( u(x) \) separately and then combine the results.

Properties of Logarithms

Understanding the properties of logarithms is crucial when working with logarithmic differentiation. These properties simplify complex expressions and make it possible to differentiate functions that would otherwise be very difficult to handle. Some key logarithmic properties include:

• Product Rule: \( {ab} = {a} + {b} \)
• Quotient Rule: \( {\frac{a}{b}} = {a} - {b} \)
• Power Rule: \( {a^b} = b {a} \)

In our exercise example, we use the Power Rule to simplify the expression \( {u(x)^{v(x)}} \) to \( v(x){u(x)} \). These properties transform the original function into a form that is more amenable to differentiation.

Differentiation Rules

The rules of differentiation are tools that help us determine the derivative of functions. We primarily use basic differentiation rules such as the power rule (derivative of \( x^n \) is \( nx^{n-1} \)), the product rule (to differentiate the product of two functions), and the quotient rule (to differentiate a division of two functions).

When encountering more complex functions, like those involving exponentials with variable bases and exponents such as \( u(x)^{v(x)} \), we often have to combine these rules for effective differentiation. In the solution for logarithmic differentiation provided, we see an application of the product rule in the derivative of the function after applying the logarithm. This results in a blend of differentiation rules to achieve the final result.