5.6: Exponential and logarithm chain rules.

Suppose an function \(u(t)\) has a derivative for all \(t\). Then

\[\left[e^{u(t)}\right]^{\prime}=e^{u(t)} u^{\prime}(t) \quad \textbf{ Exponential Chain Rule. } \label{5.34}\]

The Exponential Chain Rule is used to prove

\[[\ln t]^{\prime}=\frac{1}{t}, \quad \text{for} \quad t>0 \quad \quad \textbf{Logarithm Rule}. \label{5.35}\]

and, assuming \(u\) is positive,

\[[\ln u(t)]^{\prime}=\frac{1}{u(t)} u^{\prime}(t) \quad \quad \textbf{Logarithm Chain Rule} \label{5.36}\].

Recall Theorem 4.2.1, The Derivative Requires Continuity. If \(u(t)\) is a function and \(u ^{\prime} (t)\) exists at \(t = a\)

\[\lim _{b \rightarrow a} u(b)=u(a)\]

Proof of the exponential chain rule. We prove the rule for the case the \(u\) is an increasing function.

Suppose \(u(t)\) has a derivative at \(t = a\) and \(E(t) = e ^{u(t)}\). To simplify the argument we assume that \(u(t)\) is increasing. That is, if \(a < b\), then \(u(a) < u(b)\), and, in particular, \(u(b) - u(a) \neq 0\).

Then

\[\begin{aligned}
E^{\prime}(a)=\lim _{b \rightarrow a} \frac{e^{u(b)}-e^{u(a)}}{b-a} &=\lim _{b \rightarrow a} \frac{e^{u(b)}-e^{u(a)}}{u(b)-u(a)} \frac{u(b)-u(a)}{b-a} \\
&=\lim _{b \rightarrow a} \frac{e^{u(b)}-e^{u(a)}}{u(b)-u(a)} \times \lim _{b \rightarrow a} \frac{u(b)-u(a)}{b-a}=e^{u(a)} \times u^{\prime}(a)
\end{aligned}\]

End of Proof.

The limit

\[\lim _{b \rightarrow a} \frac{e^{u(b)}-e^{u(a)}}{u(b)-u(a)}=e^{u(a)}\]

requires some explanation. The graph of \(y = e ^x\) is shown in Figure \(\PageIndex{1}\). At every point, \(( x , e^{x} )\) of the graph, the slope of the tangent is \(e ^x\), and specifically at the point \(( u(a), e^{u(a)} )\) the slope is \(m = e ^{u(a)}\). The difference quotient

\[\frac{e^{u(b)}-e^{u(a)}}{u(b)-u(a)}\]

is the slope of a secant to the graph. Because \(\lim _{b \rightarrow a} u(b)=u(a)\) the slope of the secant approaches the slope of the tangent. Thus

\[\lim _{b \rightarrow a} \frac{e^{u(b)}-e^{u(a)}}{u(b)-u(a)}=e^{u(a)} .\]

Example 5.6.1 Find the derivatives of

\[P(x)=200 e^{-3 x} \quad Q(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}\]

Figure \(\PageIndex{1}\): Graph of \(y = e ^x \). Because \(u ^{\prime} (a)\) exists, \(u(b) \rightarrow u(a)\) as \(b \rightarrow a\). Therefore the slope of the secant, \(\frac{u(b)-u(a)}{b-a}\), approaches the slope of the tangent, \(e ^{u(a)}\).

Solutions:

\[\[\begin{array}\
&P^{\prime}(x)&=\left[200 e^{-3 x}\right]^{\prime} &Q^{\prime}(x)&=\left[\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}\right]^{\prime} \quad &\text { Logical Identity }\\
&&=200\left[e^{-3 x}\right]^{\prime} &&=\frac{1}{\sqrt{2 \pi}}\left[e^{-x^{2} / 2}\right]^{\prime} &\text { Constant Factor Rule }\\
&&=200 e^{-3 x}[-3 x]^{\prime} &&=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}\left[-x^{2} / 2\right]^{\prime} &\text { Exponential Chain Rule }\\
&&=200 e^{-3 x}(-3) &&=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}(-x) &\text { Constant Factor, Power Rules }
\end{array}\]\]

The derivative of \(P(x) = 200 e ^{-3 x}\) also can be computed using the \(e ^{k t}\) Rule. The \(e ^{k t}\) rule is a special case of the exponential chain rule:

\[\left[e^{k t}\right]^{\prime}=e^{k t}[k t]^{\prime}=e^{k t} k\]

The derivative of \(L(t) = \ln {t}\). The exponential chain rule \ref{5.34} is used to derive the natural logarithm rule, Equation \ref{5.35}. In order to use the formula \(\left[e^{u(t)}\right]^{\prime}=e^{u(t)} u^{\prime}(t)\) it is necessary to know that \(u ^{\prime} (t)\) exists. We want to compute \(\left[e^{\ln t}\right]^{\prime}\) and need to know that \([\ln t]^{\prime}\) exists. Our argument for this is shown in Figure \(\PageIndex{2}\). Also observe that \(L\) is increasing so that our argument for the exponential chain rule which assumed that \(u(t)\) was increasing is sufficient for this use. Knowing that \([\ln t]^{\prime}\) exists, it is easy to obtain a formula for it.

\[\left.\begin{array}{rlr}
e^{\ln t} & =t & (i) \\
{\left[e^{\ln t}\right]^{\prime}} & =[t]^{\prime} & (i i) \\
{[\ln t]^{\prime}} & =1 & (i i i) \\
{[\ln t]^{\prime}} & =1 & (i v) \\
{[\ln t]^{\prime}} & =\frac{1}{t} & \text { Algebra }
\end{array}\right\} \label{5.37}\]

Figure \(\PageIndex{1}\): Graphs of \(E(t) = e ^t\) and \(L(t) = \ln {t}\) and tangents at corresponding points. Because \(L\) is the inverse of \(E\), the graph of \(L\) is the reflection of the graph of \(E\) about the line \(y = x\). Because \(E\) has a tangent at \((a, b)\) (that is not horizontal! actually with positive slope), \(L\) has a tangent at \((b, a)\) (with positive slope). Therefore \(L ^{\prime} (b)\) exists for every positive value of \(b\).

We have now proved the natural logarithm rule, another Primary Formula:

Natural Logarithm Rule

\[[\ln t]^{\prime}=\frac{1}{t} \label{5.38}\]

Using the natural logarithm rule and some properties of logarithms we can differentiate \(y(t)=\ln 5 t+\ln t^{3}\):

\[\begin{aligned}
y^{\prime}(t) &=\left[\ln 5 t+\ln t^{3}\right]^{\prime}  & \text{Symbolic identity}\\
&=[\ln 5+\ln t+3 \ln t]^{\prime} & \text{Logarithm Prop's}\\
&=[\ln 5]^{\prime}+[\ln t]^{\prime}+[3 \ln t]^{\prime}  & \text{Sum Rule}\\
&=0+[\ln t]^{\prime}+[3 \ln t]^{\prime} & \text{Constant Rule}\\
&=0+[\ln t]^{\prime}+3[\ln t]^{\prime}  & \text{Constant Factor Rule}\\
&=0+\frac{1}{t}+3 \frac{1}{t}  & \text{Natural Logarithm Rule}\\
&=4 \frac{1}{t}&
\end{aligned}\]

Example 5.6.2 Logarithm functions to other bases have derivatives, but not as neat as \(1/t\). We compute \(L ^{\prime} (t)\) for \(L(t) = \log _{b} {t}\) where \(b > 0\) and \(b \neq 1\).

Solution.

\[\begin{array}\
{\left[\log _{b} t\right]^{\prime} } &=\left[\frac{\ln t}{\ln b}\right]^{\prime} \\
&=\frac{1}{\ln b}[\ln t]^{\prime} \\
&=\frac{1}{\ln b} \frac{1}{t}
\end{array} \label{5.39}\]

We summarize this as

\[\left[\log _{b} t\right]^{\prime}=\frac{1}{\ln b} \frac{1}{t} \label{5.40}\]

The Logarithm Chain Rule. We now use the Exponential Chain Rule \ref{5.34} to show that if \(u(t)\) is a positive increasing function and \(u ^{\prime} (t)\) and \([\ln u(t)]^{\prime}\) exist for all \(t\), then

\[[\ln u(t)]^{\prime}=\frac{1}{u(t)}[u(t)]^{\prime}\]

Proof. By the Exponential Chain Rule and \(e^{\ln u(t)}=u(t)\)

\[\begin{aligned}
{\left[e^{\ln u(t)}\right]^{\prime} } &=[u(t)]^{\prime} \\
e^{\ln u(t)}[\ln u(t)]^{\prime} &=u^{\prime}(t) \\
{[\ln u(t)]^{\prime} } &=\frac{1}{u(t)} u^{\prime}(t)
\end{aligned}\]

This is the

Logarithm Chain Rule

\[[\ln u(t)]^{\prime}=\frac{1}{u(t)} u^{\prime}(t) \label{5.41}\]

Example 5.6.3 Find the derivatives of

1. \(y=\ln \left(\sqrt{1-t^{2}}\right)\)
2. \(y=\ln \left(\frac{1-t}{1+t}\right)\)

Solutions:

\[\begin{array}\
{\left[\ln \left(\sqrt{1-t^{2}}\right)\right]^{\prime} } &=\left[\frac{1}{2} \ln \left(1-t^{2}\right)\right]^{\prime} &\text { Logarithm Property } \\
&=\frac{1}{2}\left[\ln \left(1-t^{2}\right)\right]^{\prime} & \text { Constant Factor }\\
&=\frac{1}{2} \frac{1}{1-t^{2}}\left[\left(1-t^{2}\right)\right]^{\prime} & \text{Generalized Logarithm Rule}\\
&=\frac{1}{2} \frac{1}{1-t^{2}}(-2 t) & \text{Sum, Constant, Constant Factor, and Power Rules}\\
\end{array}\]

2. \[\begin{array}\
   {\left[\ln \left(\frac{1-t}{1+t}\right)\right]^{\prime} } &=[\ln (1-t)-\ln (1+t)]^{\prime} &  \text { Logarithm Property } \\
   &=[\ln (1-t)]^{\prime}-[\ln (1+t)]^{\prime} &  \text { Sum Rule } \\
   &=\frac{1}{1-t}[1-t]^{\prime}-\frac{1}{1+t}[1+t]^{\prime} &  \text { Generalized Logarithm Rule } \\
   &=\frac{1}{1-t}(-1)-\frac{1}{1+t}(1) \quad &  \text { Sum, Constant, and Power Rules } \\
   &=\frac{-2}{1-t^{2}} & \text { Algebra }
   \end{array}\]

Example 5.6.4 Logarithmic Differentiation. Suppose we are to differentiate

\[y(t)=(t+2)(t+1)(t-1)\]

Proceeding indirectly, we first compute the derivative of the natural logarithm of \(y\).

\[\begin{aligned}
\ln (y(t)) &=\ln ((t+2)(t+1)(t-1)) & & \text { Logical Identity } \\
\ln (y(t)) &=\ln (t+2)+\ln (t+1)+\ln (t-1) & & \text { Logarithm Property } \\
{[\ln (y(t))]^{\prime} } &=[\ln (t+2)+\ln (t+1)+\ln (t-1)]^{\prime} & & \text { Logical Identity. } \\
\frac{1}{y(t)} y^{\prime}(t) &=\frac{1}{t+2}[t+2]^{\prime}+\frac{1}{t+1}[t+1]^{\prime}+\frac{1}{t-1}[t-1]^{\prime} & & \text { Logarithm chain rule. } \\
y^{\prime}(t) &=y(t)\left(\frac{1}{t+2}+\frac{1}{t+1}+\frac{1}{t-1}\right) & & {[t+C]^{\prime}=1 . } \\
y^{\prime}(t) &=(t+2)(t+1)(t-1)\left(\frac{1}{t+2}+\frac{1}{t+1}+\frac{1}{t-1}\right) & & \text { Definition of } y . \\
y^{\prime}(t) &=(t+1)(t-1)+(t+2)(t-1)+(t+2)(t+1) & & \text { Algebra. }
\end{aligned}\]

 

Exercises for Section 5.6, The exponential chain rule and the logarithm chain rule.

Exercise 5.6.1 Use one rule for each step and identify the rule to differentiate

1. \(P(t)=3 \ln t+e^{3 t}\)
2. \(P(t)=t^{2}+\ln 2 t\)
3. \(P(t)=\ln 5\)
4. \(P(t)=\ln \left(e^{2 t}\right)\)
5. \(P(t)=\ln \left(t^{2}+t\right)\)
6. \(P(t)=e^{t^{2}-t}\)
7. \(P(t)=e^{1 / x}\)
8. \(P(t)=\quad e^{\sqrt{x}}\)

Exercise 5.6.2 Compute the derivatives of

1. \(P(t)=e^{\left(t^{2}\right)}\)
2. \(P(t)=\ln \left(t^{2}\right)\)
3. \(P(t)=\left(e^{t}\right)^{2}\)
4. \(P(t)=e^{(2 \ln t)}\)
5. \(P(t)=\ln \left(e^{3 t}\right)\)
6. \(P(t)=\sqrt{e^{t}}\)
7. \(P(t)=e^{5}\)
8. \(P(t)=\ln (\sqrt{t})\)
9. \(P(t)=e^{t+1}\)

Exercise 5.6.3 Supply reasons that justify the steps (i) − (iv) in the equations \ref{5.37}.

Exercise 5.6.4 Give reasons for the steps (i) − (iii) in Equation \ref{5.39} deriving the derivative of the logarithm function \(L(t) = \log _{b} {t}\).

Exercise 5.6.5 Find a value for \(x\) for which \(P ^{\prime} (x) = 0\).

1. \(P(t)=x e^{-x}\)
2. \(P(t)=e^{-x}-e^{-2 x}\)
3. \(P(t)=x^{2} e^{-x}\)

Exercise 5.6.6 Use the logarithm chain rule to prove that for all numbers, \(n\):

\[\text{Power chain rule for all } n \quad \left[(u(t))^{n}\right]^{\prime}=n(u(t))^{n-1} u^{\prime}(t)\]

Assume that \(u\) is a positive increasing function and \(u ^{\prime} (t)\) exists.

Exercise 5.6.7 Use the logarithmic differentiation to compute \(y ^{\prime} (t)\) for

1. \(y(t)=t^{\pi}\)
2. \(y(t)=t^{e}\)
3. \(y(t)=\left(1+t^{2}\right)^{\pi}\)
4. \(y(t)=t^{3} e^{t}\)
5. \(y(t)=e^{\sin t}\)
6. \(y(t)=t^{t}\)

Exercise 5.6.8 Use the logarithmic differentiation to compute \(y ^{\prime} (t)\) for

1. \(y(t)=\frac{(t-1)(t+1)}{t-2}\)
2. \(y(t)=t e^{t}\)
3. \(y(t)=e^{-\frac{t^{2}}{2}}\)
4. \(y(t)=\sqrt{1+t^{2}}\)
5. \(y(t)=\frac{t^{2}}{t^{2}+1}\)
6. \(y(t)=2^{t}\)
7. \(y(t)=b^{t} \quad b>0\)
8. \(y(t)=\frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}\)
9. \(y(t)=\frac{\ln t}{e^{t}}\)