14.6 Some Derivative Calculations

14.46   Example (Derivative of $\exp$.) We know that

$$\ln(E(t)) = t \mbox{ for all }t \in \mbox{{\bf R}}.$$

If we differentiate both sides of this equation, we get

$${1\over E(t)} E'(t) = 1,$$

i.e.

$$E'(t) = E(t) \mbox{ for all }t\in \mbox{{\bf R}}.$$

14.47   Example (Derivative of $x^r.$) Let $r$ be any real number and let $f(x)=x^r$ for all $x \in \mbox{${\mbox{{\bf R}}}^{+}$}$. Then

$$f(x) = x^r = E(r\ln(x)),$$

so by the chain rule

$$f'(x) = E'(r\ln(x))\cdot{r\over x} = E(r\ln(x))\cdot rx^{-1} = x^r r x^{-1} = r x^{r-1}.$$

(Here I have used the result of exercise 14.37.) Thus the formula

$${d\over dx}(x^r) = rx^{r-1}$$

which we have known for quite a while for rational exponents, is actually valid for all real exponents.

14.48   Exercise (Derivative of $a^x$.) Let $a \in \mbox{${\mbox{{\bf R}}}^{+}$}$. Show that

$${d \over dx}(a^x) = a^x\ln(a)$$

for all $x\in \mbox{{\bf R}}.$

14.49   Example (Derivative of $x^x$.)

$$\begin{eqnarray*} {d \over dx} x^x &=& {d\over dx}e^{x\ln(x)} = e^{x\ln(x)}{d\o... ...ln(x))\\ &=& x^x ( x \cdot {1\over x} + \ln(x)) =x^x(1+\ln(x)). \end{eqnarray*}$$

Hence

$${d \over dx} x^x = x^x(1+\ln(x)) \mbox{ for all }x\in\mbox{${\mbox{{\bf R}}}^{+}$}.$$

14.50   Example (Derivative of $\arccos$.) Let $C:[0,\pi] \to [-1,1]$ be defined by

$$C(x) = \cos(x) \mbox{ for all }x \in [0,\pi].$$

We have

$$C'(x) = -\sin(x) < 0 \mbox{ for all }x \in (0,\pi),$$

so $C$ has an inverse function which is denoted by $\arccos$. By the inverse function theorem $\arccos$ is differentiable on $(-1,1)$. and we have

$$\cos(\arccos(t)) = C(\arccos(t)) = t \mbox{ for all }t\in [-1,1].$$

By the chain rule

$$-\sin(\arccos(t)) \arccos'(t) = 1 \mbox{ for all }t \in (-1,1).$$

Now since the sine function is positive on $(0,\pi)$ we get

$$\sin(s) = \sqrt{1-\cos^2(s)}$$

for all $s \in (0,\pi)$, so

$$\sin(\arccos(t)) = \sqrt{ 1 - (\cos(\arccos(t)))^2} = \sqrt{1-t^2} \mbox{ for all }t \in (-1,1).$$

Thus

$$\arccos'(t) ={-1\over \sin(\arccos(t))} = {-1 \over \sqrt{1-t^2}} \mbox{ for all }t\in(-1,1).$$

14.51   Exercise (Derivative of $\arcsin$.) Let

$$S(t) = \sin(t) \mbox{ for all }t \in [-\frac{\pi}{2},\frac{\pi}{2}].$$

Show that $S$ has an inverse function that is differentiable on the interior of its domain. This inverse functions is called $\arcsin$. Describe the domain of $\arcsin$, sketch the graphs of $S$ and of $\arcsin$, and show that

$$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}.$$

14.52   Example (Derivative of $\arctan$.) Let

$$T(x) = \tan(x) \mbox{ for all }x \in (-\frac{\pi}{2}, \frac{\pi}{2}).$$

Then T is continuous, and the image of $T$ is unbounded both above and below, so image($T$) = R. Also

$$T'(x) = \sec^2(x) > 0 \mbox{ for all }x \in (-\frac{\pi}{2}, \frac{\pi}{2})$$

so $T$ has an inverse function, which we denote by $\arctan$.

For all $x \in \mbox{{\bf R}}$

$$\tan(\arctan(x)) = T(\arctan(x)) = x,$$

so by the chain rule

$$\sec^2(\arctan(x))\arctan'(x) = 1 \mbox{ for all }x\in\mbox{{\bf R}}.$$

Now

$$\sec^2(t) = 1 + \tan^2(t) \mbox{ for all }t \in \mbox{{\rm dom}}{(\sec)},$$

so

$$\sec^2(\arctan(x)) = 1 + \tan^2(\arctan(x)) = 1+x^2 \mbox{ for all }x\in\mbox{{\bf R}}.$$

Thus

$$\arctan'(x) = {1\over \sec^2(\arctan(x))} = \frac{1}{1+x^2} \mbox{ for all }x \in \mbox{{\bf R}}.$$

14.53   Exercise (Derivative of arccot.) Let

$$V(x) = \cot(x) \mbox{ for all }x \in (0,\pi).$$

Show that $V$ has an inverse function arccot, and that

$$\frac{d}{dx} {\rm arccot}(x) = -\frac{1}{1+x^2}.$$

What is $\mbox{{\rm dom}}{({\rm arccot})}$? Sketch the graphs of $V$ and of arccot.

Remark The first person to give a name to the inverse trigonometric functions was Daniel Bernoulli (1700-1792) who used $AS$ for $\arcsin$ in 1729. Other early notations included arc(cos. = $x$) and ang(cos. = $x$)[15, page 175]. Many calculators and some calculus books use $\cos^{-1}$ to denote arccos. (If you use your calculator to find inverse trigonometric functions, make sure that you set the degree-radian-grad mode to radians.)

14.54   Exercise. A Calculate the derivatives of the following functions, and simplify your answers (Here $a$ is a constant.)

a)

$${ f(x) = x\sqrt{a^2-x^2} + a^2 \arcsin(\frac{x}{a})}$$.

b)

$${ g(x) = \arcsin(x) + \frac{x}{1-x^2}}$$.

c)

$${ h(x) = x \arccos(ax)-\frac{1}{a} \sqrt{1-a^2x^2}}$$.

d)

$${ k(x) = \arctan( e^x + e^{-x})}$$.

e)

$${ m(x) = x\sqrt{1-x^2} +\arcsin(x) (2x^2-1)}$$.

f)

$${ n(x) = e^{ax}(a\sin(bx) -b\cos(bx)).\mbox{ Here $a$ and $b$ are constants}}$$.

g)

$${ p(x) = e^{ax}(a^2x^2 - 2ax +2). \mbox{ Here $a$ is a constant.}}$$.

14.55   Exercise. A Let

$$l(x) = \arctan(\tan(x)).$$

Calculate the derivative of $l$. What is the domain of this function? Sketch the graph of $l$.

14.56   Exercise (Hyperbolic functions.) We define functions $\sinh$ and $\cosh$ on R by

$$\begin{eqnarray*} \cosh(x) &=& { e^x + e^{-x} \over 2} \mbox{ for all }x \in \mb... ...=& { e^x - e^{-x} \over 2} \mbox{ for all }x \in \mbox{{\bf R}}. \end{eqnarray*}$$

These functions are called the hyperbolic sine and the hyperbolic cosine respectively. Show that

$${d \over dx} \cosh(x) = \sinh(x),$$

and

$${d \over dx} \sinh(x) = \cosh(x).$$

Calculate

$${d \over dx} \Big(\cosh^2(x) - \sinh^2(x)\Big),$$

and simplify your answer as much as you can. What conclusion can you draw from your answer? Sketch the graphs of $\cosh$ and $\sinh$ on one set of coordinate axes.