--- title: Inverse Functions, Logarithms, and Exponentials module: Exponential, Logarithmic, and Inverse Functions moduleNumber: 6 lessonNumber: 1 order: 601 summary: > A one-to-one function has an inverse that reverses it, with a graph mirrored across y = x and a derivative given by the reciprocal-slope rule. The exponential e^x is its own derivative and the natural logarithm has derivative 1/x; logarithmic differentiation turns products, quotients, and variable powers into sums. topics: [Exponential, Logarithmic, and Inverse Functions] sources: - book: Stewart ref: "Ch. 6 — Inverse Functions; §6.1 Inverse Functions" - book: Stewart ref: "§6.2 Exponential Functions and Their Derivatives; §6.3 Logarithmic Functions" - book: Stewart ref: "§6.4 Derivatives of Logarithmic Functions" draft: false --- A function takes an input to an output. Sometimes the useful question runs the other way: given the output, which input produced it? For a bacteria culture whose population $N = f(t)$ is recorded hourly, the biologist may instead want the time $t$ required to reach a given level $N$. That reversed rule is the **inverse function** $f^{-1}$, and it exists only when $f$ never sends two inputs to the same output. The two families where this reversal is used most are the exponentials $b^x$ and the logarithms $\log_b x$, inverse to each other, and the derivative of an inverse gives both their derivatives from a single rule. ## One-to-one functions An inverse can only be defined when no output is claimed by two different inputs. If $g(2) = g(3) = 4$, there is no way to recover the input from the output $4$. > **Definition (One-to-one function).** A function $f$ is **one-to-one** if it > never takes the same value twice; that is, > $$ > f(x_1) \neq f(x_2) \quad \text{whenever} \quad x_1 \neq x_2. > $$ Geometrically this is a statement about horizontal lines. If some horizontal line $y = c$ met the graph at two points $x_1, x_2$, then $f(x_1) = f(x_2) = c$ with $x_1 \neq x_2$, breaking the definition. > **Horizontal Line Test.** A function is one-to-one if and only if no > horizontal line intersects its graph more than once. For example, $f(x) = x^3$ is one-to-one: distinct numbers have distinct cubes, and no horizontal line meets the cubic twice. By contrast $g(x) = x^2$ fails, since $g(-1) = g(1) = 1$. An increasing function is automatically one-to-one, and so is a decreasing one, because a strictly monotonic graph can never double back to repeat a height. $$ % caption: The Horizontal Line Test. A cubic meets every horizontal line % exactly once and is one-to-one; a parabola meets some lines twice and is not. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} \begin{scope} \draw[->, black] (-1.6,0) -- (1.6,0) node[right] {$x$}; \draw[->, black] (0,-1.5) -- (0,1.7) node[above] {$y$}; \draw[acc, very thick] plot[smooth, domain=-1.15:1.15] (\x, {\x*\x*\x}) node[above right] {$x^3$}; \draw[black, densely dashed] (-1.6,0.6) -- (1.6,0.6); \fill[black] (0.843,0.6) circle (1.7pt); \end{scope} \begin{scope}[xshift=4.4cm] \draw[->, black] (-1.6,0) -- (1.6,0) node[right] {$x$}; \draw[->, black] (0,-0.3) -- (0,1.9) node[above] {$y$}; \draw[acc, very thick] plot[smooth, domain=-1.32:1.32] (\x, {\x*\x}) node[above right] {$x^2$}; \draw[black, densely dashed] (-1.6,0.64) -- (1.6,0.64); \fill[black] (-0.8,0.64) circle (1.7pt); \fill[black] (0.8,0.64) circle (1.7pt); \end{scope} \end{tikzpicture} $$ ## The inverse function > **Definition (Inverse function).** Let $f$ be one-to-one with domain $A$ and > range $B$. Its **inverse** $f^{-1}$ has domain $B$ and range $A$, and is > defined by > $$ > f^{-1}(y) = x \iff f(x) = y. > $$ The inverse reverses the assignment: if $f$ sends $x$ to $y$, then $f^{-1}$ sends $y$ back to $x$. Domain and range trade places, $$ \operatorname{dom} f^{-1} = \operatorname{ran} f, \qquad \operatorname{ran} f^{-1} = \operatorname{dom} f. $$ The superscript $-1$ is notation, not an exponent: $f^{-1}(x)$ does not mean $1/f(x)$. Applying a function and then its inverse returns the starting value, which is the content of the **cancellation equations**: $$ f^{-1}\bigl(f(x)\bigr) = x \ \text{ for } x \in A, \qquad f\bigl(f^{-1}(x)\bigr) = x \ \text{ for } x \in B. $$ To compute a formula for $f^{-1}$, solve $y = f(x)$ for $x$, then interchange the names of the variables so the independent variable is again called $x$. > **Finding an inverse.** Write $y = f(x)$; solve for $x$ in terms of $y$; > interchange $x$ and $y$ to get $y = f^{-1}(x)$. For $f(x) = x^3 + 2$, solving $y = x^3 + 2$ gives $x = \sqrt[3]{y - 2}$, and after interchanging, $f^{-1}(x) = \sqrt[3]{x - 2}$. The rule "cube, then add 2" is reversed by "subtract 2, then take the cube root." The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$. The point $(a, b)$ lies on the graph of $f$ exactly when $(b, a)$ lies on the graph of $f^{-1}$, and $(b, a)$ is the mirror image of $(a, b)$ in the diagonal. $$ % caption: A one-to-one function and its inverse are reflections of each other % across the line y = x; the point (a,b) on f maps to (b,a) on the inverse. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-0.3,0) -- (4.6,0) node[right] {$x$}; \draw[->, black] (0,-0.3) -- (0,4.6) node[above] {$y$}; \draw[black, dashed] (-0.2,-0.2) -- (4.4,4.4) node[right, black] {$y=x$}; % f: increasing curve above the diagonal \draw[acc, very thick] plot[smooth, domain=0.35:2.05] (\x, {0.55*\x*\x*\x - 0.2*\x + 0.35}) node[right] {$f$}; % f^{-1}: reflection (swap coordinates) \draw[acc, very thick, densely dashed] plot[smooth, domain=0.35:2.05] ({0.55*\x*\x*\x - 0.2*\x + 0.35}, \x) node[above] {$f^{-1}$}; % matched points \fill[black] (1.6,{0.55*1.6*1.6*1.6 - 0.2*1.6 + 0.35}) circle (2pt); \node[anchor=south east] at (1.6,{0.55*1.6*1.6*1.6 - 0.2*1.6 + 0.35}) {$(a,\,b)$}; \fill[black] ({0.55*1.6*1.6*1.6 - 0.2*1.6 + 0.35},1.6) circle (2pt); \node[anchor=north west] at ({0.55*1.6*1.6*1.6 - 0.2*1.6 + 0.35},1.6) {$(b,\,a)$}; \draw[black] (1.6,{0.55*1.6*1.6*1.6 - 0.2*1.6 + 0.35}) -- ({0.55*1.6*1.6*1.6 - 0.2*1.6 + 0.35},1.6); \end{tikzpicture} $$ A function that fails the Horizontal Line Test can often be salvaged by restricting its domain. The full parabola $y = x^2$ has no inverse, but $f(x) = x^2$ on $[0, 2]$ is one-to-one, with inverse $f^{-1}(x) = \sqrt{x}$ on $[0, 4]$. ## The derivative of an inverse Reflecting a graph across $y = x$ interchanges the roles of the horizontal and vertical directions, so it interchanges rise and run. The slope of $f^{-1}$ at a point is therefore the reciprocal of the slope of $f$ at the mirror point. > **Theorem (Derivative of an inverse).** If $f$ is one-to-one and > differentiable with $f'\bigl(f^{-1}(a)\bigr) \neq 0$, then $f^{-1}$ is > differentiable at $a$ and > $$ > \bigl(f^{-1}\bigr)'(a) = \frac{1}{f'\bigl(f^{-1}(a)\bigr)}. > $$ The reason is [implicit differentiation](/calculus/derivatives/implicit-differentiation-and-related-rates). Write $y = f^{-1}(x)$, so $f(y) = x$. Differentiating $f(y) = x$ with respect to $x$ and using the [chain rule](/calculus/derivatives/differentiation-rules-and-the-chain-rule), $$ f'(y)\,\frac{\d y}{\d x} = 1 \qquad\Longrightarrow\qquad \frac{\d y}{\d x} = \frac{1}{f'(y)} = \frac{1}{\d x/\d y}. $$ In Leibniz notation the rule reads $\dfrac{\d y}{\d x} = \dfrac{1}{\,\d x/\d y\,}$, which is the reciprocal-slope statement written symbolically. $$ % caption: Writing g = f^{-1}, reflection across y = x sends the tangent to f % at (a,b), of slope m, to the tangent to g at (b,a), of reciprocal slope 1/m. \begin{tikzpicture}[scale=1.05, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-0.3,0) -- (3.7,0) node[right] {$x$}; \draw[->, black] (0,-0.3) -- (0,3.7) node[above] {$y$}; \draw[black, dashed] (-0.2,-0.2) -- (3.5,3.5) node[right, black] {$y=x$}; \draw[acc, very thick] plot[smooth, domain=0.35:3.35] (\x, {0.25*\x*\x}) node[right] {$f$}; \draw[acc, very thick, densely dashed] plot[smooth, domain=0.35:3.35] ({0.25*\x*\x}, \x) node[above] {$g$}; % tangent to f at P=(2.4,1.44), slope 1.2 \draw[black] (1.75,{1.44-1.2*0.65}) -- (3.05,{1.44+1.2*0.65}); \fill[black] (2.4,1.44) circle (1.7pt) node[anchor=north west] {$(a,\,b)$}; \node[black, anchor=north west, font=\footnotesize] at (2.55,0.78) {slope $m$}; % tangent to f^{-1} at P'=(1.44,2.4), slope 5/6 \draw[black, densely dashed] ({1.44-0.78},{2.4-0.833*0.78}) -- ({1.44+1.4},{2.4+0.833*1.4}); \fill[black] (1.44,2.4) circle (1.7pt) node[anchor=south east] {$(b,\,a)$}; \node[black, anchor=south east, font=\footnotesize] at (0.72,2.72) {slope $\frac{1}{m}$}; \end{tikzpicture} $$ The formula gives the derivative of an inverse without ever computing the inverse itself. > **Worked example.** For $f(x) = 2x + \cos x$, find $(f^{-1})'(1)$. > > The derivative $f'(x) = 2 - \sin x$ is positive everywhere, so $f$ increases > and is one-to-one, and the inverse exists. Rather than solve $2x + \cos x = 1$ > for a formula, read off the point directly: $f(0) = 2(0) + \cos 0 = 1$, so > $f^{-1}(1) = 0$. Then > > $$ > \bigl(f^{-1}\bigr)'(1) = \frac{1}{f'\bigl(f^{-1}(1)\bigr)} = \frac{1}{f'(0)} > = \frac{1}{2 - \sin 0} = \frac{1}{2}. > $$ ## Exponential functions An **exponential function** has the form $f(x) = b^x$ with a positive constant base $b$; the variable sits in the exponent. This is not the power function $x^b$, where the variable is the base. For a positive integer $n$, $b^n$ is $b$ multiplied by itself $n$ times; the laws of exponents extend the definition to rational $x$, and a limiting process fills the irrational values so that $b^x$ is continuous everywhere. > **Theorem (Exponential function).** If $b > 0$ and $b \neq 1$, then > $f(x) = b^x$ is continuous with domain $\mathbb{R}$ and range $(0, \infty)$, > so $b^x > 0$ for all $x$. It is increasing if $b > 1$ and decreasing if > $0 < b < 1$. For $a, b > 0$ and real $x, y$: > $$ > b^{x+y} = b^x b^y, \qquad > b^{x-y} = \frac{b^x}{b^y}, \qquad > (b^x)^y = b^{xy}, \qquad > (ab)^x = a^x b^x. > $$ Every graph $y = b^x$ passes through $(0, 1)$ because $b^0 = 1$, and larger bases climb more steeply for $x > 0$. The end behavior follows from the monotonicity: $$ b > 1: \quad \lim_{x \to \infty} b^x = \infty, \quad \lim_{x \to -\infty} b^x = 0; \qquad 0 < b < 1: \quad \lim_{x \to \infty} b^x = 0, \quad \lim_{x \to -\infty} b^x = \infty. $$ In either case the $x$-axis is a horizontal asymptote. $$ % caption: The family y = b^x. Every curve passes through (0,1); bases above 1 % increase and bases below 1 decrease, each a reflection of another across the % vertical axis. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-2.6,0) -- (2.7,0) node[right] {$x$}; \draw[->, black] (0,-0.3) -- (0,3.7) node[above] {$y$}; \draw[acc, very thick] plot[smooth, domain=-2.5:1.65] (\x, {pow(3,\x)}) node[right] {$3^x$}; \draw[acc, very thick] plot[smooth, domain=-2.5:1.85] (\x, {pow(2,\x)}) node[right] {$2^x$}; \draw[acc, very thick, densely dashed] plot[smooth, domain=-1.85:2.5] (\x, {pow(0.5,\x)}) node[above left] {$2^{-x}$}; \fill[black] (0,1) circle (2pt) node[anchor=north east] {$(0,\,1)$}; \draw[black, dotted] (-2.6,0) -- (2.7,0); \end{tikzpicture} $$ ### The number e and the derivative of the exponential Differentiating $b^x$ from the definition exposes a constant that depends only on the base: $$ f'(x) = \lim_{h \to 0} \frac{b^{x+h} - b^x}{h} = b^x \lim_{h \to 0} \frac{b^h - 1}{h} = f'(0)\, b^x. $$ The rate of change of an exponential is proportional to its own value, with proportionality constant $f'(0)$, the slope at the $y$-intercept. Numerically $f'(0) \approx 0.69$ for $b = 2$ and $\approx 1.10$ for $b = 3$, so some base between $2$ and $3$ gives slope exactly $1$. That base is the number $e$. > **Definition (The number e).** $e$ is the base for which > $$ > \lim_{h \to 0} \frac{e^h - 1}{h} = 1, > $$ > equivalently the base whose graph $y = e^x$ crosses the $y$-axis with slope > $1$. To twenty places, $e \approx 2.71828182845904523536$, and $e$ is > irrational. With $f'(0) = 1$, the proportionality formula collapses to the defining property of the natural exponential function: it is its own derivative. > **Theorem (Derivative of $e^x$).** > $$ > \frac{\d}{\d x}\, e^x = e^x, > \qquad\text{and with the chain rule,}\qquad > \frac{\d}{\d x}\, e^{u} = e^{u}\,\frac{\d u}{\d x}. > $$ The slope of the tangent to $y = e^x$ at any point equals the height of the curve there. No other elementary function reproduces itself under differentiation, which is why $e$ rather than $10$ or $2$ is the natural base for calculus. $$ % caption: For y = e^x the slope of the tangent at each point equals the % y-coordinate at that point; here the tangent at (1, e) has slope e. \begin{tikzpicture}[scale=1.0, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-2.3,0) -- (2.3,0) node[right] {$x$}; \draw[->, black] (0,-0.3) -- (0,4.4) node[above] {$y$}; \draw[acc, very thick] plot[smooth, domain=-2.2:1.45] (\x, {pow(2.71828,\x)}) node[above left] {$y=e^x$}; % point (1, e); e ~ 2.718 \fill[black] (1,2.718) circle (2pt); \node[anchor=west] at (1.05,2.55) {$(1,\,e)$}; % tangent line at (1,e): slope e, y = e*x \draw[black, thick] (0.1,{2.718*0.1}) -- (1.55,{2.718*1.55}); \node[black, anchor=north west] at (1.35,{2.718*1.35}) {slope $=e$}; \end{tikzpicture} $$ For example, $\dfrac{\d}{\d x}\, e^{\tan x} = e^{\tan x}\sec^2 x$ by the chain rule, and the product rule gives $\dfrac{\d}{\d x}\bigl(e^{-4x}\sin 5x\bigr) = e^{-4x}(5\cos 5x - 4\sin 5x)$. The integral counterpart is $\displaystyle\int e^x\,\d x = e^x + C$. ## Logarithmic functions Since $b^x$ is one-to-one for $b \neq 1$, it has an inverse, the **logarithm to base $b$**. > **Definition (Logarithm).** For $b > 0$, $b \neq 1$, > $$ > \log_b x = y \iff b^y = x. > $$ > Thus $\log_b x$ is the exponent to which $b$ must be raised to produce $x$. The cancellation equations become $$ \log_b(b^x) = x \ \text{ for all } x \in \mathbb{R}, \qquad b^{\log_b x} = x \ \text{ for all } x > 0. $$ As the inverse of the exponential, $\log_b$ has domain $(0, \infty)$, range $\mathbb{R}$, and a graph obtained by reflecting $y = b^x$ across $y = x$. Every logarithmic graph passes through $(1, 0)$ because $\log_b 1 = 0$. The rapid growth of $b^x$ reflects into the very slow growth of $\log_b x$. The properties of exponents transpose into the properties of logarithms. > **Theorem (Laws of logarithms).** For $b > 1$, $\log_b$ is one-to-one, > continuous, and increasing on $(0, \infty)$ with range $\mathbb{R}$. For > $x, y > 0$ and any real $r$: > $$ > \log_b(xy) = \log_b x + \log_b y, \quad > \log_b\!\left(\tfrac{x}{y}\right) = \log_b x - \log_b y, \quad > \log_b(x^r) = r\log_b x. > $$ The vertical asymptote of the logarithm is the reflected image of the exponential's horizontal asymptote: $$ \lim_{x \to \infty} \log_b x = \infty, \qquad \lim_{x \to 0^+} \log_b x = -\infty \qquad (b > 1). $$ ### The natural logarithm The most convenient base is $e$; its logarithm gets the special notation $\ln x = \log_e x$, the **natural logarithm**. The defining relations and cancellation equations specialize to $$ \ln x = y \iff e^y = x, \qquad \ln(e^x) = x, \qquad e^{\ln x} = x, \qquad \ln e = 1. $$ These two cancellation identities solve equations with an unknown in the exponent. > **Worked example.** Solve $e^{5 - 3x} = 10$ for $x$. > > The unknown sits in the exponent, so apply the inverse of $e^{(\cdot)}$: take > $\ln$ of both sides and use $\ln(e^{5 - 3x}) = 5 - 3x$. > > $$ > 5 - 3x = \ln 10 > \qquad\Longrightarrow\qquad > x = \tfrac{1}{3}(5 - \ln 10) \approx 0.8991. > $$ Any logarithm can be rewritten in terms of $\ln$, which is why calculators need only one logarithm key. > **Theorem (Change of base).** For any base $b > 0$, $b \neq 1$, > $$ > \log_b x = \frac{\ln x}{\ln b}. > $$ The proof is one line: if $y = \log_b x$ then $b^y = x$; taking $\ln$ gives $y \ln b = \ln x$, so $y = \ln x / \ln b$. As an application, $\log_8 5 = \ln 5 / \ln 8 \approx 0.773976$. $$ % caption: The natural exponential and natural logarithm are inverses, so their % graphs mirror across y = x. Because e^x has slope 1 at (0,1), ln x has slope 1 % at (1,0). \begin{tikzpicture}[scale=0.95, >=stealth, font=\small] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-2.6,0) -- (3.4,0) node[right] {$x$}; \draw[->, black] (0,-2.6) -- (0,3.4) node[above] {$y$}; \draw[black, dashed] (-2.4,-2.4) -- (3.2,3.2) node[right, black] {$y=x$}; \draw[acc, very thick] plot[smooth, domain=-2.4:1.2] (\x, {pow(2.71828,\x)}) node[above] {$y=e^x$}; \draw[acc, very thick, densely dashed] plot[smooth, domain=0.09:3.2] (\x, {ln(\x)}) node[right] {$y=\ln x$}; \fill[black] (0,1) circle (1.8pt); \fill[black] (1,0) circle (1.8pt); \end{tikzpicture} $$ ## Derivatives of logarithms The natural logarithm is differentiable because it inverts the differentiable function $e^x$. Writing $y = \ln x$, so $e^y = x$, and differentiating implicitly gives $e^y\,\d y/\d x = 1$, hence $\d y/\d x = 1/e^y = 1/x$. > **Theorem (Derivative of $\ln$).** > $$ > \frac{\d}{\d x}\,\ln x = \frac{1}{x}, > \qquad > \frac{\d}{\d x}\,\ln u = \frac{1}{u}\,\frac{\d u}{\d x} = \frac{u'}{u}. > $$ The chain-rule form is the one used in practice. For $y = \ln(x^3 + 1)$, $$ \frac{\d y}{\d x} = \frac{1}{x^3 + 1}\cdot 3x^2 = \frac{3x^2}{x^3 + 1}, $$ and $\dfrac{\d}{\d x}\ln(\sin x) = \dfrac{\cos x}{\sin x} = \cot x$. Extending to absolute values gives a formula valid on both sides of the origin, $$ \frac{\d}{\d x}\ln|x| = \frac{1}{x}, \qquad\text{whose integral form}\qquad \int \frac{1}{x}\,\d x = \ln|x| + C $$ fills the one gap in the power rule for antiderivatives: $\int x^n\,\d x$ gives $x^{n+1}/(n+1)$ except when $n = -1$, and that missing case is supplied by the logarithm. As one consequence, $\int \tan x\,\d x = \ln|\sec x| + C$, found by substituting $u = \cos x$. The general and base-$b$ formulas follow from $\ln$ through change of base and the identity $e^{\ln b} = b$. | Function | Derivative | Source | | --- | --- | --- | | $e^x$ | $e^x$ | defining property | | $b^x$ | $b^x \ln b$ | $b^x = e^{(\ln b)x}$ | | $\ln x$ | $1/x$ | inverse of $e^x$ | | $\log_b x$ | $1/(x \ln b)$ | change of base | The base-$b$ exponential derivative comes from rewriting $b^x = e^{(\ln b)x}$ and applying the chain rule: $$ \frac{\d}{\d x}\,b^x = \frac{\d}{\d x}\,e^{(\ln b)x} = e^{(\ln b)x}\cdot \ln b = b^x \ln b. $$ The base-$b$ logarithm derivative is $\dfrac{\d}{\d x}\log_b x = \dfrac{1}{x \ln b}$, smallest and simplest when $b = e$ so that $\ln b = 1$. That simplicity is the reason calculus prefers natural logarithms. ## Logarithmic differentiation Products, quotients, and powers stacked together are painful to differentiate directly, but taking a logarithm first turns them into sums and differences. The technique is **logarithmic differentiation**. > **Logarithmic differentiation.** To differentiate $y = f(x)$: take $\ln$ of > both sides and simplify with the laws of logarithms; differentiate implicitly; > solve for $y'$; substitute the expression for $y$. > **Worked example.** Differentiate $y = \dfrac{x^{3/4}\sqrt{x^2 + 1}}{(3x + 2)^5}$. > > Taking logarithms separates the three factors into a sum: > > $$ > \ln y = \tfrac{3}{4}\ln x + \tfrac{1}{2}\ln(x^2 + 1) - 5\ln(3x + 2). > $$ > > Differentiate implicitly, one term at a time: > > $$ > \frac{y'}{y} = \frac{3}{4x} + \frac{x}{x^2 + 1} - \frac{15}{3x + 2}. > $$ > > Multiplying by $y$ gives the derivative without a single application of the > quotient rule: > > $$ > y' = \frac{x^{3/4}\sqrt{x^2 + 1}}{(3x + 2)^5} > \left(\frac{3}{4x} + \frac{x}{x^2 + 1} - \frac{15}{3x + 2}\right). > $$ The same device proves the **power rule** for an arbitrary real exponent. Let $y = x^n$ with $x \neq 0$; then $\ln|y| = n\ln|x|$, so $y'/y = n/x$ and $y' = n\,y/x = n x^{n-1}$. Functions of the form $[g(x)]^{h(x)}$, where both base and exponent vary, need the same treatment or the equivalent rewrite $e^{h(x)\ln g(x)}$; neither the power rule nor the exponential rule applies on its own. > **Worked example.** Differentiate $y = x^{\sqrt{x}}$ for $x > 0$. > > Both base and exponent hold the variable, so neither $\dfrac{\d}{\d x}x^n$ nor > $\dfrac{\d}{\d x}b^x$ applies. Take $\ln$ of both sides: > > $$ > \ln y = \sqrt{x}\,\ln x. > $$ > > Differentiate implicitly, using the product rule on the right: > > $$ > \frac{y'}{y} = \frac{1}{2\sqrt{x}}\,\ln x + \sqrt{x}\cdot\frac{1}{x} > = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}. > $$ > > Solving for $y'$ and restoring $y = x^{\sqrt{x}}$, > > $$ > y' = x^{\sqrt{x}}\left(\frac{\ln x + 2}{2\sqrt{x}}\right). > $$ ## The number e as a limit The derivative $\ln'(1) = 1$ can be unwound into a limit formula for $e$ itself. From the definition of the derivative at $x = 1$, $$ 1 = \ln'(1) = \lim_{x \to 0}\frac{\ln(1 + x) - \ln 1}{x} = \lim_{x \to 0}\ln(1 + x)^{1/x}. $$ By continuity of the exponential, exponentiating both sides gives $$ e = \lim_{x \to 0}(1 + x)^{1/x}, \qquad\text{equivalently}\qquad e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}. $$ The second form, with $n = 1/x$, is the expression that reappears in [continuously compounded growth](/calculus/exponential-logarithmic-and-inverse-functions/growth-decay-inverse-trig-and-hyperbolic-functions): letting the compounding interval shrink to zero drives the accumulation factor to $e$.