Several Differentiation Drills

For any integer $n$ , you may freely use the results

$\begin{aligned} (x^n)' &= nx^{n-1},\\ \sin'(x) &= \cos(x),\\ \cos'(x) &= -{\sin(x)},\\ \tan'(x) &= \sec^2(x), \\ (e^x)' &= e^x, \\ \ln'(x) &= \frac 1x, \end{aligned}$

from either previous posts or from real analysis. You may also apply the chain rule, product rule, and quotient rule however you wish.

Problem 1. For any $a > 0$ , prove that

$\displaystyle \begin{aligned} (a^x)' = a^x \ln(a), \quad \log_a'(x) = \frac{1}{x \ln(a)}. \end{aligned}$

(Click for Solution)

Solution. By exponential properties,

$a^x = (e^{\ln(a)})^x = e^{x \ln(a)}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (a^x)' &= (e^{x \ln(a)})'\\ &= e^{x \ln(a)} \cdot \frac{\mathrm d}{\mathrm dx}(x \ln a) \\ &= e^{x \ln(a)} \cdot \ln(a) \\ &= a^x \ln(a). \end{aligned}$

By logarithm properties and linearity,

$\displaystyle \begin{aligned} \log_a'(x) &= \left( \frac{\ln(x)}{\ln(a)} \right)'\\ &= \frac{1}{\ln(a)} \ln'(x) \\ &= \frac{1}{\ln(a)} \cdot \frac 1x = \frac 1{x \ln (a)}. \end{aligned}$

Up till now, we have proven that $(x^n)' = nx^{n-1}$ only for integer values of $n$ . We extend this result to all real values of $n$ .

Problem 2. For any real $r$ , use Problem 1 to prove that $(x^r)' = rx^{r-1}$ .

(Click for Solution)

Solution. By exponential properties,

$x^r = (e^{\ln x})^r = e^{r \ln x}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (x^r)' &= (e^{r \ln x})'\\ &= e^{r \ln x} \cdot \frac{\mathrm d}{\mathrm dx}(r \ln x) \\ &= x^r \cdot r \cdot \frac 1x = rx^{r-1}. \end{aligned}$

Example 1. By Problem 2,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{\sqrt 2}) = \sqrt 2 \cdot x^{\sqrt 2 - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^\pi) = \pi \cdot x^{\pi - 1}.$

In particular,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}\neq e^x = \frac{\mathrm d}{\mathrm dx}(e^x).$

Problem 3. Prove that

$\displaystyle \sec'(x) = \sec(x)\tan(x),\quad \csc'(x) = -{\csc(x)\cot(x)}.$

(Click for Solution)

Solution. Writing $\displaystyle \sec(x) = \frac 1{\cos(x)}$ , apply the quotient rule to obtain

$\displaystyle \begin{aligned} \sec'(x) &= \frac{ (1)' \cdot \cos(x) - 1 \cdot \cos'(x) }{ \cos^2(x) } \\ &= \frac{ 0 \cdot \cos(x) - 1 \cdot (-\sin(x)) }{ \cos^2(x) } \\ &= \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x) \tan(x). \end{aligned}$

We recall that $\csc(x) = \sec(\pi/2-x)$ so that

$\displaystyle \begin{aligned}\csc'(x) &= (\sec(\pi/2-x))' \\ &= \sec'(\pi/2-x)) \cdot (-1) \\ &= \sec(\pi/2-x) \tan(\pi/2-x) \cdot (-1) \\ &= \csc(x) \cot(x) \cdot (-1) \\ &= - \csc(x) \cot(x). \end{aligned}$

Problem 4. Given that $t > 0$ , $x \equiv x(t) > 0$ , and $y \equiv y(t) > 0$ , evaluate $\displaystyle \frac{\mathrm d}{\mathrm dt}(x^y)$ .

(Click for Solution)

Solution. Writing $x^y = e^{y \ln(x)}$ and using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dt}(x^y) &= \frac{\mathrm d}{\mathrm dt} (e^{y \ln(x)}) \\ &= e^{y \ln(x)} \left( \frac{\mathrm d}{\mathrm dt}(y \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac{\mathrm d}{\mathrm dt} ( \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac 1x \cdot \frac{\mathrm dx}{\mathrm dt} \right) \\ &= x^y \ln(x) \cdot \frac{\mathrm dy}{\mathrm dt} + yx^{y-1} \cdot \frac{\mathrm dx}{\mathrm dt}. \end{aligned}$

Remark 1. If $x = t$ and $y$ is constant, we recover the power rule. If $x$ is constant and $y = t$ , we recover Problem 1. Furthermore, if $y = x = t$ , we obtain

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x) = x^x (\ln(x) + 1).$

In the language of partial derivatives, denoting $z = x^y \equiv { x(t) }^{ y(t) }$ ,

$\displaystyle \frac{\mathrm d z}{\mathrm dt} = \frac{\partial z}{\partial y} \cdot \frac{\mathrm dy}{\mathrm dt} + \frac{\partial z}{\partial x} \cdot \frac{\mathrm dx}{\mathrm dt}.$

Problem 5. Suppose $f, f^{-1}$ are differentiable on $\mathbb R$ . Evaluate $(f^{-1})'(x)$ .

(Click for Solution)

Solution. By definition of the inverse function,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} (x) &= \frac{\mathrm d}{\mathrm dx} (f(f^{-1}(x))) \\ 1 &= f'(f^{-1}(x)) \cdot \frac{\mathrm d}{\mathrm dx}(f^{-1}(x)) \\ &= f'(f^{-1}(x)) \cdot f'(f^{-1}(x)) \\ (f^{-1})'(x) &= \frac 1{f'(f^{-1}(x))}. \end{aligned}$

—Joel Kindiak, 9 Feb 25, 2126H

KindiakMath

Several Differentiation Drills

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Several Differentiation Drills

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