# 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.