Mixed Differentiation Practice

This web exercise gives practice applying differentiation rules to a variety of differentiation problems.

All the calculus steps are shown in the solutions, but only minimal algebraic simplication is done.

If you want practice with the formulas only, then study Differentiation Formula Practice.

There are two versions of this exercise—you may want to look at both to see which works best for you. The other version has much stricter parenthesis usage.

The Differentiation Rules

The differentiation formulas used in these practice problems are compiled in the earlier lesson Differentiation Formula Practice. Only one new technique is needed, which is discussed next.

Differentiating a Variable Base to a Variable Power

An expression like $\,x^{2x}\,$ has both a variable base and variable exponent.

This is different from power functions, which have only a variable base, like $\,x^2\,.$ It is also different from exponential functions, which have only a variable exponent , like $\,2^x\,.$

To differentiate a problem of the form

$${\text{(variable base)}}^{\text{(variable power)}}$$

you need what is sometimes called the ‘log trick’. The technique is illustrated next.

Differentiate: $y = x^{2x}$

Preparation for Solution

We already know how to differentiate $\,{\text{e}}^{f(x)}\,$:

$$ \frac{d}{dx}{\text{e}}^{f(x)} = {\text{e}}^{f(x)}\cdot f'(x)\tag{1} $$

Is it possible to rename $\,x^{2x}\,$ so that it has a base of $\,\text{e}\,,$ and something we know how to differentiate in the exponent? Yes! Keep reading!

Here are the tools needed in the renaming process:

First Tool: Inverse Relationship Between $\,\ln x\,$ and $\,{\text{e}}^x$

Because of the inverse relationship between $\,\ln x\,$ and $\,{\text{e}}^x\,,$ we have:

$${\text{e}}^{\ln x} = x\ \ \ \text{for all } x \gt 0$$

We'll use this ‘backwards’:

$$x = {\text{e}}^{\ln x}\ \ \ \text{for all } x \gt 0\tag{2}$$

Equation (2) says that you can take any positive ‘stuff ’ and write it as $\,{\text{e}}^{\ln{(\text{stuff})}}\,.$ Although this is a more complicated name, it's precisely the name needed in this situation!

Second Tool: A Property of Logarithms

Recall this property of logarithms:

$$\ln x^y = y\ln x\tag{3}$$

It says, roughly, that you can bring exponents down. It holds for values of $\,x\,$ and $\,y\,$ for which both $\,x \gt 0\,$ and $\,x^y\gt 0\,.$

The ‘Log Trick’ for Differentiating $\,x^{2x}$

Now, we're ready to differentiate $\,x^{2x}\,.$

To avoid undefinedness problems, we only consider $\,x^{2x}\,$ for $\,x \gt 0\,.$ As long as $\,x \gt 0\,,$ then $\,x^{2x}\,$ is defined and positive.

Step 1: Rename $\,x^{2x}$

$$ \begin{align} y &= x^{2x}\cr &\quad \text{(original name)}\cr\cr &= {\text{e}}^{\ln x^{2x}}\cr &\quad \text{(use equation (2))}\cr\cr &= \color{red}{{\text{e}}^{2x\ln x}}\cr &\quad \text{(use equation (3))} \end{align} $$

This is now a name of the form $\,\color{red}{{\text{e}}^{f(x)}}\,,$ where we know how to differentiate $\,f(x)\,$!

Step 2: Differentiate Using the New Name

$$ \begin{gather} y = {\text{e}}^{2x\ln x}\cr \text{(the new name)} \end{gather} $$
$$ \begin{align} \frac{dy}{dx} &= {\text{e}}^{2x\ln x}\cdot \frac{d}{dx} (2x\ln x)\cr &\qquad \text{(use equation (1))}\vphantom{\Rule{1pt}{1.3em}{1pt}}\cr\cr &= \color{blue}{{\text{e}}^{2x\ln x}} \color{orange}{\bigl(2x\cdot\frac{1}{x} + (\ln x)\cdot 2\bigr)}\cr &\qquad \text{(use the Product Rule)}\vphantom{\Rule{1pt}{1.3em}{1pt}} \end{align} $$

Step 3: As Much as Possible, Rename Derivative to Match Original Function

It's always recommended to rename the derivative in a way that matches the original function, as much as possible.

Reversing earlier steps, we can write:

$$ \color{blue}{{\text{e}}^{2x\ln x} = {\text{e}}^{\ln x^{2x}} = x^{2x}} $$

Also:

$$ \begin{align} &\color{orange}{2x\cdot \frac 1x + (\ln x)\cdot 2}\cr\cr &\qquad =\color{orange}{2 + 2\ln x}\cr\cr &\qquad =\color{orange}{2(1 + \ln x)} \end{align} $$

Combining results:

$$\frac{d}{dx} x^{2x} = 2x^{2x}(1 + \ln x)$$

This same ‘log trick’ works for differentiating other functions with both variable bases and variable powers.

In the exercises below, the calculus steps in finding the derivatives are shown. However, only minimal simplifications are done:

Constants are often multiplied together and pulled to the front
Rational exponents are written as radicals

Your teacher may want a different ‘name’ for the answer than the one given here!