Question

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

Short answer

Question: Find the derivative of $$x^{10x}$$ with respect to x. Answer: The derivative of $$x^{10x}$$ with respect to x is $$\frac{d}{d x}\left(x^{10 x}\right) = x^{10x}\left(10\ln(x) + 10\right)$$.

Step-by-step solution

Apply Natural Logarithm on both the sides

Take the natural logarithm of both sides of the equation $$y = x^{10x}$$. Using the logarithmic properties, we can rewrite the equation as follows: $$\ln(y) = \ln\left(x^{10x}\right)$$ $$\ln(y) = 10x \ln(x)$$

Implicit Differentiation with respect to x

Differentiate both sides of the equation $$\ln(y) = 10x \ln(x)$$ with respect to x. Remember to use the differentiation rules and consider the chain rule when differentiating the left side. $$\frac{d}{dx}\left(\ln(y)\right) = \frac{d}{dx}\left(10x \ln(x)\right)$$ Using the chain rule on the left side and the product rule on the right side, we get: $$\frac{1}{y}\frac{dy}{dx} = 10\ln(x) + 10x\cdot\frac{1}{x}$$

Solve for dy/dx

Multiply both sides by y to isolate $$\frac{dy}{dx}$$ and then substitute $$y = x^{10x}$$ back. $$\frac{dy}{dx} = y\left(10\ln(x) + 10x\cdot\frac{1}{x}\right)$$ $$\frac{dy}{dx} = x^{10x}\left(10\ln(x) + 10\right)$$ The derivative of $$x^{10x}$$ with respect to x is: $$\frac{d}{d x}\left(x^{10 x}\right) = x^{10x}\left(10\ln(x) + 10\right)$$

Key concepts

Derivative of Exponential Functions

Exponential functions are those where the variable appears in the exponent. The standard form is usually written as \(f(x) = a^{x}\) or more complex variants like the one in our problem, \(x^{10x}\).
To differentiate such functions, especially non-standard forms, logarithmic differentiation is very useful. This technique simplifies the function by applying natural logs and using logarithmic identities.
In our case, given the function \(y = x^{10x}\), taking the natural log transforms it into a multiplicative form:
\(\ln(y) = 10x \ln(x)\).
This new form is easier to differentiate.
Exponential differentiation often requires combining other derivative techniques like the product rule, particularly when the base is not a constant, but a function of \(x\).

Implicit Differentiation

In implicit differentiation, we differentiate equations that are not explicitly solved for one variable in terms of another.
Here, our function in logarithmic form, \(\ln(y) = 10x \ln(x)\), is set up for implicit differentiation since \(y\) is a function of \(x\).
We differentiate both sides with respect to \(x\).
On the left, the chain rule applies because \(y\) is an implicit function of \(x\), leading to \(\frac{1}{y}\frac{dy}{dx}\).
The right side requires the product rule, since it is a product of \(10x\) and \(\ln(x)\).
Implicit differentiation is extremely useful for such situations where \(y\) is not isolated, allowing us to solve simultaneously for \(\frac{dy}{dx}\).

Chain Rule

The chain rule is a fundamental technique in differentiation used to differentiate composite functions.
Consider a function \(g(x) = f(u(x))\). To differentiate \(g\) with respect to \(x\), you first differentiate \(f\) with respect to \(u\), \(f'(u)\), and then multiply by the derivative of \(u\) with respect to \(x\), \(u'(x)\):
\(\frac{d}{dx}g(x) = f'(u) \cdot u'(x)\).
In the logarithmic differentiation process, on the left side when differentiating \(\ln(y)\), the chain rule is used to handle \(y\), which is implicitly a function of \(x\), resulting in \(\frac{1}{y}\frac{dy}{dx}\).
The chain rule is indispensable for correctly solving such derivatives, ensuring that each layer of a function's composition is accounted for.