Tag: Difference Quotient

Determine whether given statements about a function differentiable at a point are true or false

by RoRi

September 19, 2015

Let $f$ be a function and $a$ a point such that $f'(a)$ exists. Determine which of the following are true or false.

$\displaystyle{f'(a) = \lim_{h \to a} \frac{f(h) - f(a)}{h-a}}$ .
$\displaystyle{f'(a) = \lim_{h \to 0} \frac{f(a) - f(a -h)}{h}}$ .
$\displaystyle{ f'(a) = \lim_{t \to 0} \frac{f(a+2t) - f(a)}{t}}$ .
$\displaystyle{ f'(a) = \lim_{t \to 0} \frac{f(a+2t) - f(a+t)}{2t}}$ .

True. Let $k = h-a$ . Then $k \to 0$ as $(h-a) \to 0$ implies $k \to 0$ as $h \to a$ . Thus,
$f'(a) = \lim_{k \to 0} \frac{f(a+k) - f(a)}{k} = \lim_{h \to a} \frac{f(a+h-a) - f(a)}{h-a} = \lim_{h \to a} \frac{f(h) - f(a)}{h-a}. \qquad \blacksquare$
True. Let $k = -h$ , then
$\lim_{h \to 0} \frac{f(a) - f(a-h)}{h} = \lim_{k \to 0} \frac{f(a) - f(a+k)}{-k} = \lim_{k \to 0} \frac{f(a+k) - f(a)}{k} = f'(a). \qquad \blacksquare$
False. First, from the definition of limit we can conclude
$\lim_{t \to 0} f(t) = \lim_{2t \to 0} f(t)$

(by taking $\delta' = \frac{\delta}{2}$ ). Then we have,

$\begin{align*} \lim_{t \to 0} \frac{f(a+2t) - f(a)}{t} &= \lim_{2t \to 0} \frac{f(a+2t) - f(a)}{t} \\ &= 2 \lim_{2t \to 0} \frac{f(a+2t) - f(a)}{2t} \\ &= 2 f'(a) \neq f'(a) & \text{if } f'(a) \neq 0. \end{align*}$

Thus, this is not true in general (since there are many function differentiable at a point $a$ such that the derivative at that point is nonzero). ( Note: This statement would be true were the denominator in the given equation $2t$ instead of $t$ .)
False. Here we evaluate,
$\begin{align*} \lim_{t \to 0} \frac{f(a+2t) - f(a+t)}{2t} &= \frac{1}{2} \left( \lim_{t \to 0} \frac{f(a+2t)-f(a)}{t} - \lim_{t \to 0} \frac{f(a+t) - f(a)}{t} \right) \\ &= \frac{1}{2} ( 2f'(a) - f'(a)) \qquad (\text{part (c)})\\ &= \frac{1}{2} f'(a) \neq f'(a) \qquad \text{if } f'(a) \neq 0. \end{align*}$

Find values of constants so that the derivative of a function exists

by RoRi

September 18, 2015

Consider the function $f$ defined by

$f(x) = \begin{cases} \sin x & \text{if } x \leq c, \\ ax+b & \text{if } x > c. \end{cases}$

Find values for the constants $a$ and $b$ such that the derivative $f'(c)$ exists.

We know that $f'(c)$ exists if and only if

$\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$

exists. This limit exists if and only if the two one-sided limits exits and are equal:

$\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}.$

Using the definition of $f$ , we then evaluate these limits,

$\begin{align*} && \lim_{h \to 0^+} \frac{a(c+h)+ b - \sin c}{h} &= \lim_{h \to 0^-} \frac{\sin (c+h) - \sin c}{h} \\ \implies && \lim_{h \to 0^+} \left(\frac{ac+b- \sin c}{h}\right) + a &= \lim_{h \to 0^-} \frac{\sin c \cos h + \sin h \cos c - \sin c}{h} \\ \implies && \lim_{ h \to 0^+} \left(\frac{ac+b - \sin c}{h}\right) + a &= \lim_{h \to 0^-}\left( \frac{\sin c \cdot \left( \cos h - 1 \right)}{h} \right) + \cos c. \end{align*}$

In simplifying the right hand side we used that $\lim_{h \to 0} \frac{\sin h}{h} = 1$ . Furthermore, for the limit on the left to exist we must have $ac+b - \sin c = 0$ (otherwise the limit will diverge as $h \to 0$ ). Now for the expression on the right, we claim the limit in the expression is 0. We can see this because

$\begin{align*} \lim_{h \to 0^-} \frac{\sin c (\cos h - 1)}{h} &= \sin c \lim_{h \to 0^-} \frac{\cos h - 1}{h} \\ &= \sin c \lim_{h \to 0^-} \frac{\cos (0 + h) - \cos 0}{h} \end{align*}$

But this limit is the derivative of $\cos x$ at $x = 0$ . Since $(\cos x)' = -\sin x$ and $\sin 0 = 0$ , this term is indeed 0.
So, coming back to our original equations we then have,

$\lim_{ h \to 0^+} \left(\frac{ac+b - \sin c}{h}\right) + a &= \lim_{h \to 0^-}\left( \frac{\sin \left( \cos h - 1 \right)}{h} \right) + \cos c \quad \implies \quad a = \cos c.$

Furthermore, since we already established that $ac + b - \sin c = 0$ we have,

$b = \sin c - c \cos c.$

Therefore, the expressions for $a$ and $b$ we are asked to find are,

$a = \cos c, \qquad b = \sin c - c \cos c.$

Find values of constants so that the derivative of a function exists

by RoRi

September 18, 2015

Consider the function $f$ defined by

$f(x) = \begin{cases} \frac{1}{|x|} & \text{if } |x| > c, \\ a+bx^2 & \text{if } |x| \leq c. \end{cases}$

Find values for the constants $a$ and $b$ such that the derivative $f'(c)$ exists.

First, let’s assume $c > 0$ otherwise $f(x) = \frac{1}{|x|}$ everywhere and the values of $a$ and $b$ are arbitrary (since the value of $f$ does not depend on them).

Then, $f'(c)$ exists implies the one-sided limits exist and are equal:

$\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}.$

Using the definition of $f$ we plug in the expressions for $f$ when approaches from the left and from the right,

$\begin{align*} &&\lim_{h \to 0^+} \frac{\frac{1}{c+h} - a - bc^2}{h} &= \lim_{h \to 0^-} \frac{a+b(c+h)^2 - a - bc^2}{h} \\ \implies && \lim_{h \to 0^+} \frac{1-(a+bc^2)(c+h)}{h(c+h)} &= 2bc \\ \implies && \lim_{h \to 0^+} \frac{1 - ac - ah - bc^3 - bc^2h}{h(c+h)} &= 2bc \\ \implies && \lim_{h \to 0^+} \left( \frac{1 - ac - bc^3}{h(c+h)} \right) - \frac{a+bc^2}{c} &= 2bc. \end{align*}$

The limit on the left exists if and only if $1-ac -bc^3 = 0$ (otherwise the limit diverges to infinity has $h \to 0$ ). In this case the limit is 0, and we have two equations,

$1 - ac - bc^3 = 0 \qquad \text{and} \qquad - \frac{a + bc^2}{c} = 2bc.$

Using the equation on the right we have, $a = -3bc^2$ . Then plugging this value into the equation on the left we obtain

$1 + 3bc^3 - bc^3 = 0 \quad \implies \quad b = - \frac{1}{2c^3}.$

Finally, using this in our expression $a = -3bc^2$ we solve for $a$ in terms of $c$ alone, $a = \frac{3}{2c}$ . Therefore,

$a = \frac{3}{2c}, \qquad b = - \frac{1}{2c^3}.$

Find values of constants so that the derivative of a function exists

by RoRi

September 18, 2015

Consider the function $f$ defined by

$f(x) = \begin{cases} x^2 & \text{if } x \leq c, \\ ax+b & \text{if } x > c. \end{cases}$

Find values for the constants $a$ and $b$ such that the derivative $f'(c)$ exists.

We know that the derivative $f'(c)$ exists if and only if

$\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$

exists. Furthermore, this limit exists if and only if the one-sided limits both exist and are equal:

$\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}.$

So, plugging in the formula for $f$ (which is $ax + b$ if we approach $c$ from the right, and is $x^2$ if we approach from the left, and noting that $f(c) = c^2$ from the definition of $f$ ) we have,

$\begin{align*} &&\lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} &= \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} \\ \implies && \lim_{h \to 0^+} \frac{a(c+h) + b - c^2}{h} &= \lim_{h \to 0^-} \frac{(c+h)^2 - c^2}{h} \\ \implies && \lim_{h \to 0^+} \frac{ac + b - c^2}{h} + a &= 2c. \end{align*}$

For the limit on the left to exist we must have $ac + b - c^2 = 0$ (otherwise the limit will diverge as $h \to 0$ ). Furthermore, this limit must be 0 since $ac +b - c^2$ is a constant (and the limit of $0/h$ as $h \to 0$ is 0). Therefore, we have $a = 2c$ , and we have the equation

$ac + b - c^2 = 0 \quad \implies \quad 2c^2 + b - c^2 = 0 \quad \implies \quad b = -c^2.$

Therefore, $a = 2c$ and $b = -c^2$ are the values of the requested constants in terms of $c$ .

Use the binomial theorem to prove the formula for the derivative of x^n

by RoRi

September 16, 2015

Consider the function $f(x) = x^n$ for a positive integer $n$ . Prove that

$\frac{f(x+h) - f(x)}{h} = nx^{n-1} + \frac{n(n-1)}{2} x^{n-2} h + \cdots + nxh^{n-2} + h^{n-1}.$

Conclude that $f'(x) = nx^{n-1}$ by considering the limit of this as $h \to 0$ .

Proof. We recall the binomial theorem,

$(x+h)^n = \sum_{k=0}^n \binom{n}{k} x^k h^{n-k}.$

So, then we calculate,

$\begin{align*} \frac{f(x+h) - f(x)}{h} &= \frac{1}{h} \left( (x+h)^n - x^n \right) \\ &= \frac{1}{h} \left( \left(\sum_{k=0}^n \binom{n}{k} x^k h^{n-k} \right) - x^n \right) \\ &= \frac{1}{h} \left( \sum_{k=0}^{n-1} \binom{n}{k} x^k h^{n-k} \right) \\ &= \sum_{k=0}^{n-1} \binom{n}{k} x^k h^{n-k-1} \\ &= nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} h + \cdots + h^{n-1}. \qquad \blacksquare \end{align*}$

Taking the limit as $h \to 0$ of both sides of the equation we conclude,

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = nx^{n-1}.$

Evaluate the limit

by RoRi

August 24, 2015

Evaluate

$\lim_{h \to 0} \frac{(t+h)^2 - t^2}{h}.$

First, we simplify the expression,

$\begin{align*} \frac{(t+h)^2 - t^2}{h} &= \frac{t^2 + 2th + h^2 - t^2}{h} \\ &= \frac{2th + h^2}{h} \\ &= 2t + h. \end{align*}$

Then, we evaluate the limit,

$\lim_{h \to 0} (2t + h) = 2t.$

Prove difference quotient formulas for sine and cosine

by RoRi

August 2, 2015

For $h \neq 0$ prove

$\begin{align*} \frac{\sin (x+h) - \sin x}{h} &= \frac{\sin \left(\frac{h}{2}\right)}{h/2} \cos \left(x + \frac{h}{2} \right) \\ \phantom{=} \\ \frac{\cos (x+h) - \cos x}{h} &= - \frac{\sin \left( \frac{h}{2} \right)}{h/2} \sin \left(x + \frac{h}{2} \right). \end{align*}$

Proof. We use the sine and cosine of sum and difference formulas in the following computations:

$\begin{align*} \frac{\sin (x+h) - \sin x}{h} &= \frac{ \sin \left( x + \frac{h}{2} + \frac{h}{2} \right) - \sin \left( x + \frac{h}{2} - \frac{h}{2} \right)}{h} \\ &= \frac{\sin \left(x + \frac{h}{2} \right) \cos \left( \frac{h}{2} \right) + \sin \left( \frac{h}{2} \right) \cos \left( x + \frac{h}{2} \right) - \sin \left(x + \frac{h}{2} \right) \cos \left(\frac{h}{2} \right) + \sin \left( \frac{h}{2} \right) \cos \left( x + \frac{h}{2} \right)}{h} \\ &= \frac{2 \sin \frac{h}{2} }{h} \cos \left(x + \frac{h}{2} \right) \\ &= \frac{\sin \frac{h}{2}}{h/2} \cos \left(x + \frac{h}{2} \right). \end{align*}$

And,

$\begin{align*} \frac{ \cos (x+h) - \cos x}{h} &= \frac{ \cos \left( x + \frac{h}{2} + \frac{h}{2} \right) - \cos \left(x + \frac{h}{2} - \frac{h}{2} \right)}{h} \\ &= \frac{ \cos \left(x + \frac{h}{2} \right) \cos \frac{h}{2} - \sin \left( x + \frac{h}{2} \right) \sin \frac{h}{2} - \cos \left(x + \frac{h}{2} \right) \cos \frac{h}{2} - \sin \left(x + \frac{h}{2} \right) \sin \frac{h}{2}}{h}\\ &= - \frac{2 \sin \left( x + \frac{h}{2} \right) \sin \frac{h}{2}}{h} \\ &= - \frac{ \sin \frac{h}{2}}{h/2} \sin \left( x + \frac{h}{2} \right). \qquad \blacksquare \end{align*}$