# Transcendental Functions - Examples

On this page we display several examples to illustrate the concepts and theorems discussed in the Transcendental Functions theme.

## Differentiating an inverse function

**Problem**

Show that the function \(f(x) = x \sqrt{3+x^2}\) has an inverse, and compute \((f^{-1})'(-2)\).

**Solution**

We have that \( f'(x) = \sqrt{3+x^2} + \frac{x^2}{\sqrt{3+x^2}} \), which is positive for every \(x\). \(f\) is therefore increasing, therefore also one-to-one, and therefore has an inverse function \(f^{-1}\). We then know that

\[ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}.\]

We see that to compute the \((f^{-1})'(-2)\), we first need to find \(f^{-1}(-2)\). Now, this is defined as the value of \(y\) for which \(f(y) = y \sqrt{3+y^2} = -2 \). This is easily found using Maple, but we can also find it analytically:

\[ \begin{align} y \sqrt{3+y^2} = -2 \Longrightarrow & y^2(3+y^2) = 4 & \\ \Longleftrightarrow & y^4+3y^2-4 = 0 & (\mathrm{second \ degree \ equation \ in \ } y^2) \\ \Longleftrightarrow & y^2 = 1 \quad \mathrm{or} \quad y^2 = -4 & \end{align} \] Since a square is not negative, we conclude that \(y^2 = 1\), and examining the original equation, we conclude that \(y = -1\). We easily find \(f'(-1) = 5/2\), and thus \((f^{-1})'(-2) = 2/5\).

## Manipulating logarithms

**Problem**

Solve \(2 \log_3x+\log_9x = 10\).

**Solution**

Since \(f(x) = 3^x\) is one-to-one, we can apply \(f\) to both sides of the equation without changing the solution set.
\[3^{2 \log_3x+\log_9x} = 3^{10} \]
\[3^{2\log_3x}3^{\log_9x} = 3^{10}\]
\[(3^{\log_3x})^2 (9^{\log_9x})^{\frac12} = 3^{10} \]
\[x^2x^{\frac12} = 3^{10}\]
\[x = 3^{10 \cdot \frac25} \]
We conclude that \(x = 81\).

## Proving an identity with logarithms

**Problem**

Let \(x, a, b >0\), with \(a, b \neq 1\). Prove that \[\log_a x = \frac{\log_b x}{\log_b a}.\]

**Solution**
Let \(u = \log_a x\). This means that \(x = a^u\). Taking the logarithm with base \(b\) on both sides yelds \(\log_b a^u = \log_b x\). Simplifying this, we get \(u\cdot\log_b a = \log_b x\), and after dividing by \(\log_b a \) on both sides: \(u = \log_a x = \frac{\log_b x}{\log_b a}\), and the formula is proved.

## Logarithmic differentiation

**Problem**

Find \(F'(0)\), where
\[ F(x) = \frac{ (x^2+7)\sqrt{3+2x}(1-x)^{1/3}}{(1+5x)^{4/5} }. \]

**Solution**

Taking the logarithm of a product transforms it into a sum. Sums are easier to differentiate than products. So let's consider \(\ln(F(x))\) (allowed since \(F(x)>0\) when \(x\) is near \(0\)):

\[ \ln(F(x)) = \ln(x^2+7) + \frac12 \ln(3+2x) + \frac13 \ln(1-x) - \frac{4}{5} \ln (1+5x). \]

This function can easily be differentiated (implicitly):

\[ \frac{F'(x)}{F(x)} = \frac{2x}{x^2+7} + \frac12 \frac{2}{3+2x} + \frac13 \frac{-1}{1-x} - \frac{4}{5} \frac{1}{1+5x}. \]

In other words we get \[ F'(x) = \left( \frac{2x}{x^2+7} + \frac12 \frac{2}{3+2x} + \frac13 \frac{-1}{1-x} - \frac{4}{5} \frac{1}{1+5x} \right) \frac{ (x^2+7)\sqrt{3+2x}(1-x)^{1/3}}{(1+5x)^{4/5} }.\]

## Exponential cooling

**Problem**

Find the temperature of an object 5 minutes after it is removed from an oven at 72 degrees and placed outdoors in 20 degrees, given that the object's temperature after 1 minute is 48 degrees.

**Solution**

We know from Newton's law of cooling that the temperature \(T(t)\) satisfies the equation \[\frac{\mathrm{d}T}{\mathrm{d}t}=k(T-20).\] Let \(u(t) = T(t) -20\). Then we have that \(u(0)=72-20 =52\) and \(u(1) =28\). Furthermore, \(u\) satisfies \[\frac{\mathrm{d}u}{\mathrm{d}t}=ku,\] with solution \(u(t) =u(0)\mathrm{e}^{kt} =52\mathrm{e}^{kt}.\) To find the constant \(k\), note that \(28=u(1) =52\mathrm{e}^{k},\) so that \(\mathrm{e}^{k} = 28/52 =7/13,\) giving \(k =\ln(7/13).\)

We want to find \(T(5) =u(5)+20\), and from the calculations above we have \[u(5)=52\mathrm{e}^{5\ln(7/13)}=2.35,\] so that the temperature after five minutes is \(2.35+20 = 22.35\) degrees.