Question

Evaluate the following derivatives. $$\frac{d}{d x}\left((2 x)^{4 x}\right)$$

Short answer

Answer: The derivative of the expression \((2x)^{4x}\) with respect to \(x\) is \((2x)^{4x}(4\ln(2x) + 2)\).

Step-by-step solution

Rewrite the given expression using natural logarithm

We will first rewrite the expression using natural logarithm: $$y = (2x)^{4x}$$ Taking the natural logarithm of both sides we have: $$\ln(y) = \ln((2x)^{4x})$$ Now, apply the property of logarithms that allow us to move the exponent to the front: $$\ln(y) = 4x\ln(2x)$$

Apply implicit differentiation to the new equation

Now we will differentiate both sides of the equation with respect to \(x\) using implicit differentiation: $$\frac{d}{dx}\ln(y) = \frac{d}{dx}(4x\ln(2x))$$ On the left-hand side, we will apply the chain rule, differentiating the outer function \(\ln(y)\) first with respect to \(y\) and then multiply it by the derivative of the inner function \(y\) with respect to \(x\). On the right-hand side, we will apply the product rule: $$\frac{1}{y}\frac{dy}{dx} = 4\ln(2x) + 4x\left(\frac{1}{2x}\right)$$ Notice that \(\frac{d}{dx}\ln(2x)\) requires the chain rule as well: $$\frac{1}{y}\frac{dy}{dx} = 4\ln(2x) + \frac{4x}{2x}$$

Simplify and isolate the derivative on one side

Now we will simplify the expression and isolate the derivative \(\frac{dy}{dx}\) on one side of the equation: $$\frac{1}{y}\frac{dy}{dx} = 4\ln(2x) + 2$$ So, we have: $$\frac{dy}{dx} = y(4\ln(2x) + 2)$$

Replace y with the original function

Since we know \(y = (2x)^{4x}\), we will substitute this back into the equation to get the final answer: $$\frac{dy}{dx} = (2x)^{4x}(4\ln(2x) + 2)$$

Key concepts

Implicit Differentiation

Implicit differentiation is a clever technique used when you have an equation where the dependent and independent variables are mixed together, and it's tough to express one variable explicitly in terms of the other.
When a function is given implicitly, it means it’s not solved for the dependent variable. For example, in our case, after taking the logarithm we ended up with the equation \( \ln(y) = 4x \ln(2x) \), rather than having \(y\) isolated on one side.
To differentiate implicitly, follow these steps:

• Differentiating both sides of the equation with respect to the independent variable.
• Use derivative rules like the product, chain, and power rules as necessary.
• Don’t forget to multiply by \( \frac{dy}{dx} \) when differentiating the dependent variable terms.

This technique is handy when dealing with natural logarithms or more complicated functions where direct differentiation is not feasible.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with base \(e\), where \(e\approx 2.718\). It is exceptionally useful in calculus because of its nice derivative properties. The natural logarithm can help simplify functions for easier differentiation.
In the original exercise, we used it to handle the exponential form \((2x)^{4x}\). By taking the natural logarithm, we were able to bring the exponent down in front of the logarithm, thus making it possible to apply differentiation rules more manageably.

• Using \( \ln(a^b) = b\ln(a) \), you can rewrite complex exponential expressions.
• This step transforms the problem into a more familiar form for applying calculus techniques.
• After differentiation, remember to revert back using the properties that helped you convert initially.

Natural logarithms simplify exponents and are an invaluable part of implicit differentiation.

Product Rule

The product rule is a differentiation rule used when a function is expressed as the product of two other functions. It's necessary because simply differentiating each part separately may lead to the wrong result.
The product rule states: if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is\[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]In our solution, after taking the logarithm and differentiating, the equation \(4x \ln(2x)\) appears. Here, \(4x\) and \(\ln(2x)\) are the two functions being multiplied, thus the product rule is applied.

• First, differentiate \(4x\), giving 4.
• Next, differentiate \(\ln(2x)\), using the chain rule, which we'll discuss next.
• Finally, apply the formula: \(4 \ln(2x) + 4x\frac{1}{2x} \).

Recognizing when to use the product rule is essential for correct differentiation.

Chain Rule

The chain rule is crucial when differentiating composite functions. This rule allows us to differentiate expressions where one function is "inside" another by differentiating the outer function and multiplying it by the derivative of the inner function.
The chain rule can be expressed as: if you have a composite function \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\).
In our example, differentiating \(\ln(2x)\) requires using the chain rule since 2x is "inside" the logarithmic function. We first differentiate the \(\ln \) function to get \( \frac{1}{2x} \), then multiply by the derivative of \(2x\), which is 2.

• Differentiating \(\ln(2x)\) gives \( \frac{1}{2x} \times 2 \).
• This simplifies to 1, as used in the solution.
• Identifying nested functions quickly can make applying this rule easier.

Without the chain rule, problems like these would be much more complex to solve effectively.