Derivatives of logarithmic, exponential and inverse trigonometric functions

This week the derivative table is completed. Once the derivatives of exponential, logarithmic and inverse trigonometric functions combine with the chain rule, you can differentiate every function in Calculus I. Some of these are pure memorization; the places that really earn points are logarithmic differentiation and the derivative of an inverse function. First a short list:

1. The derivative of $e^x$ is itself. For $e^{u}$ it is $e^u\cdot u'$.
2. $(\ln u)'=\dfrac{u'}{u}$. Before differentiating, break the expression apart with log rules.
3. When the exponent is a variable ($x^x$) or there are many factors, take $\ln y$, then differentiate.
4. In inverse trigonometric derivatives, a root or $1+u^2$ is in the denominator. For the chain rule, $u'$ goes in the numerator.
5. $(f^{-1})'(b)=\dfrac{1}{f'(a)}$, where $f(a)=b$.

Exponential functions

The number $e$ is the base chosen so that the slope of the tangent to $e^x$ at $x=0$ is exactly $1$. Therefore:

$\frac{d}{dx}e^x=e^x$

The slope of the function at every point equals its own value. For another base, write $a^x=e^{x\ln a}$ and apply the chain rule:

$\frac{d}{dx}a^x=a^x\ln a$

Example 1. $y=e^{x^2-3x}$: $y'=(2x-3)e^{x^2-3x}$. The exponent stays as it is, and the derivative of the exponent goes in front.

Example 2. $y=x^2e^{-x}$: $y'=2xe^{-x}+x^2(-e^{-x})=e^{-x}(2x-x^2)$. After the product rule, factor out $e^{-x}$; in week 9 the sign analysis is easy in this form.

Example 3. $y=2^{\sin x}$: $y'=2^{\sin x}\ln 2\cdot\cos x$.

PITFALL · applying the power rule to an exponential. $(2^x)'=x\cdot2^{x-1}$ is wrong. The power rule holds when the base is the variable and the exponent is constant ($x^2$). When the base is constant and the exponent is a variable ($2^x$), use the exponential rule: $2^x\ln2$.

Logarithmic functions

If $y=\ln x$, then $e^y=x$. Implicit derivative: $e^yy'=1 \Rightarrow y'=\dfrac{1}{e^y}=\dfrac1x$.

$\frac{d}{dx}\ln x=\frac1x,\qquad \frac{d}{dx}\ln|x|=\frac1x\ (x\ne0),\qquad \frac{d}{dx}\log_a x=\frac{1}{x\ln a}$

The most used form is the one with the chain rule: $\dfrac{d}{dx}\ln u=\dfrac{u'}{u}$.

Example 4. $y=\ln(\sin x)$: $y'=\dfrac{\cos x}{\sin x}=\cot x$.

Example 5 (break it apart first). $y=\ln\dfrac{x^3\sqrt{x+1}}{(2x-1)^2}$.

Don’t bother with the quotient rule. Expand first with log rules:

$y=3\ln x+\tfrac12\ln(x+1)-2\ln(2x-1)$

Then term by term:

$y'=\frac3x+\frac{1}{2(x+1)}-\frac{4}{2x-1}$

In the last term the $2$ came from the chain rule: $-2\cdot\dfrac{2}{2x-1}$.

Logarithmic differentiation

Use it in three situations:

• The base and the exponent are both variable: $x^x$, $x^{\sin x}$, $(\ln x)^x$.
• Many factors in the numerator and denominator: $\dfrac{(x^2+1)^3\sqrt{x-1}}{(x+2)^4}$.
• Roots and powers nested together.

Algorithm:

1. ln y = ln(right side)        take ln of both sides
2. expand with log rules         exponents to the front, products to sums
3. differentiate both sides      left side: y'/y
4. y' = y · (derivative of the right side)   write y' with the original formula

Example 6. $y=x^{\sin x}$, $x>0$.

$\ln y=\sin x\ln x \Rightarrow \frac{y'}{y}=\cos x\ln x+\frac{\sin x}{x} \Rightarrow y'=x^{\sin x}\left(\cos x\ln x+\frac{\sin x}{x}\right)$

PITFALL · choosing one of the two rules for $x^x$. The power rule gives $x\cdot x^{x-1}=x^x$, the exponential rule gives $x^x\ln x$. Both are wrong; the correct answer is the sum of the two: $x^x(1+\ln x)$. This is not a coincidence, logarithmic differentiation accounts for both variabilities.

The derivative of an inverse function

If $f$ is one-to-one and differentiable, $f(a)=b$ and $f'(a)\ne0$:

$(f^{-1})'(b)=\frac{1}{f'(a)}$

Graphically: if the tangent slope of $f$ at the point $(a,b)$ is $m$, then under reflection in $y=x$ the slope at the point $(b,a)$ is $\tfrac1m$.

Example 7. $(f^{-1})'(2)$ for $f(x)=x^3+x$.

f(a) = 2  ->  a^3 + a = 2  ->  a = 1      (by trial)
f'(x) = 3x^2 + 1  ->  f'(1) = 4
(f^-1)'(2) = 1/4

You don’t need to find the formula of the inverse. This is the route the exam expects.

Inverse trigonometric functions

If $y=\arcsin x$, then $\sin y=x$ and $y\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$. Implicit derivative:

$\cos y\cdot y'=1 \Rightarrow y'=\frac{1}{\cos y}=\frac{1}{\sqrt{1-\sin^2y}}=\frac{1}{\sqrt{1-x^2}}$

The sign of the root is $+$, because $\cos y\ge0$ on that interval. The others come out the same way:

The derivatives of $\arcsin$ and $\arccos$ differ only in sign. This is a consequence of $\arcsin x+\arccos x=\tfrac{\pi}{2}$ being constant.

Example 8. $y=\arctan\sqrt x$: $y'=\dfrac{1}{1+(\sqrt x)^2}\cdot\dfrac{1}{2\sqrt x}=\dfrac{1}{2\sqrt x(1+x)}$.

Example 9. $y=\arcsin 2x$: $y'=\dfrac{2}{\sqrt{1-4x^2}}$.

The real importance of these derivatives shows up in weeks 12-14: the integrals $\displaystyle\int\frac{dx}{1+x^2}=\arctan x+C$ and $\displaystyle\int\frac{dx}{\sqrt{1-x^2}}=\arcsin x+C$ are this table read backwards.

Mixed examples: which rule when?

On exams functions don’t come alone; three rules work together inside one question. Ask yourself these questions in order:

Example 10. $y=e^{-2x}\arctan(x^2)$.

The outer structure is a product. The derivative of the first factor is $-2e^{-2x}$ (chain rule). The derivative of the second is $\dfrac{2x}{1+x^4}$ (chain rule, $u=x^2$, $u^2=x^4$). Combine:

$y'=-2e^{-2x}\arctan(x^2)+e^{-2x}\cdot\frac{2x}{1+x^4}=2e^{-2x}\left(\frac{x}{1+x^4}-\arctan(x^2)\right)$

Example 11. $y=\ln\big(x+\sqrt{x^2+1}\big)$.

$y'=\frac{1+\dfrac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{\dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+1}}$

An expression that looks long reduces to a single term. This result will come back in week 14 as the answer to the integral $\displaystyle\int\frac{dx}{\sqrt{x^2+1}}$.

Why does $\ln|x|$ matter?

$\ln x$ is defined only for $x>0$. $\ln|x|$ is defined everywhere $x\ne0$, and its derivative is $\tfrac1x$ on both sides: for $x<0$, $\ln|x|=\ln(-x)$, and by the chain rule its derivative is $\dfrac{-1}{-x}=\dfrac1x$. That is why, when integrating, you write $\displaystyle\int\frac{dx}{x}=\ln|x|+C$; forgetting the absolute value leaves the answer undefined for negative $x$.

Exponential growth

For $y=Ce^{kt}$, $y'=kCe^{kt}=ky$. That is, the rate of change is proportional to the current amount. Population, interest and radioactive decay are all modeled with this equation. If you write the “doubles every 3 hours” model from week 1 with base $e$, you get $k=\dfrac{\ln2}{3}$.

The limit definition of e

The derivative of $\ln x$ at $x=1$ is $1$. From the definition:

$1=\lim_{h\to0}\frac{\ln(1+h)}{h}=\lim_{h\to0}\ln(1+h)^{1/h}$

Since $e^x$ is continuous, exponentiate both sides: $\displaystyle\lim_{h\to0}(1+h)^{1/h}=e$. This is the continuous version of the limit $(1+\tfrac1n)^n\to e$ you saw in week 1.

Common mistakes