4.1 Arithmetic of derivatives

Look at the Sum rule

True

Try an example, like $f(x)=g(x)=x$.

False, in general

Simplify

True

$g(x)=f(x)+f(x)+f(x)$

If you're creative, you can find lots of ways to differentiate!
Constant multiple: $g'(x)=3f'(x)$.
Product rule: $g'(x) = \diff{}{x}\{3\}f(x)+3f'(x)=0f(x)+3f'(x)=3f'(x)$.
Sum rule: $g'(x)=\diff{}{x}\{f(x)+f(x)+f(x)\}=f'(x)+f'(x)+f'(x)=3f'(x)$.
Quotient rule: $g'(x)=\diff{}{x}\left\{\frac{f(x)}{\frac{1}{3}}\right\}= \frac{\frac{1}{3}f'(x)-f(x)(0)}{\frac{1}{9}}=\frac{\frac{1}{3}f'(x)}{\frac{1}{9}}=9\left(\frac{1}{3}\right)f'(x)=3f'(x)$.
All rules give $g'(x)=3f'(x)$.

Use linearity and the known derivatives of $x^2$ and $x^{1/2}$.

$f'(x)=6x+\frac{2}{\sqrt{x}}$

Remember $\ds\diff{}{x}\{2^x\}=2^x\log2$.

$f^{(n)}=2^x(\log 2)^n$

You have already seen $\diff{}{x}\{\sqrt{x}\}$.

$-36x+24\sqrt{x}+\frac{20}{\sqrt{x}}-45$

The equation of a line can be determined using a point, and the slope. The derivative of $x^3$ can be found by writing $x^3=(x)(x^2)$.

$y - \frac{1}{8} = \frac{3}{4}\cdot \left(x-\frac{1}{2}\right)$, or $y= \tfrac{3}{4} x - \tfrac{1}{4}$

Be careful to distinguish between speed and velocity.

(a) $4$ (b) left (c) decreasing

$\dfrac{1}{{(x+1/2)}^2}$, or $\dfrac{4}{(2x+1)^2}$

How do you take care of that power?

$-72$

After you differentiate, factor out $e^x$.

$g'(x)=[f(x)+f'(x)]e^x$

Population growth is rate of change of population.

$b'(t)-d'(t)$

$(1,3),\ (3,27)$

Interpret it as a derivative that you know how to compute.

$\dfrac{1}{2\sqrt{100180}}$

The answer is not 10 square metres per second.

$20t+7$ square metres per second.

You don't need to know $g(0)$ or $g'(0)$.

0

First expression, $f(x)=\dfrac{g(x)}{h(x)}$:

Second expresson, $f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}$:

and this is exactly what we got from differentiating the first expression.

In order to be differentiable, a function should be continuous. To determine the differentiability of the function at $x=1$, use the definition of the derivative.

$a=b=\dfrac{e}{2}$

Review Pascal's Triangle.

(a) $g''(x)=[f(x)+2f'(x)+f''(x)]e^x$
(b) $g'''(x)=[f(x)+3f'(x)+3f''(x)+f'''(x)]e^x$
(c) $g^{(4)}(x)=[f(x)+4f'(x)+6f''(x)+4f'''(x)+f^{(4)}(x)]e^x$

Check signs

In the quotient rule, there is a minus, not a plus. Also, $2(x+1)+2x$ is not the same as $2(x+1)$.

The correct version is:

Read Lemma 4.1.14 carefully.

False

Quotient rule

$\dfrac{(x-1)e^x}{2x^2}$

$e^{2x}=\left(e^x\right)^2$

$2e^{2x}$

$e^{a+x}=e^ae^x$

$e^{a+x}$

Figure out where the derivative is positive.

$x>-1$

The acceleration is given by $s''(t)$.

$0$

Product rule will work nicely here. Alternately, review the result of Question 25.

$2e^{2x}$

To find the sign of a product, compare the signs of each factor. The function $e^t$ is always positive.

When $t$ is in the interval $(-2,0)$.

Use factorials, as in Example 3.4.2.

$\dfrac{3}{15!}$

First, factor an $x$ out of the derivative. What's left over looks like a quadratic equation, if you take $x^2$ to be your variable, instead of $x$.

$4x(x^2+2)(x^2+3)$

$\frac{1}{t}=t^{-1}$

$12t^3+15t^2+\frac{1}{t^2}$

First simplify. Don't be confused by the role reversal of $x$ and $y$: $x$ is just the name of the function $\big(2y+\tfrac{1}{y}\big)\cdot y^3$, which is a function of the variable $y$. You are to differentiate with respect to $y$.

$x'(y)=8y^3+2y$

$\sqrt{x}=x^{1/2}$

$T'(x)=\dfrac{(x^2+3)\left(\frac{1}{2\sqrt{x}}\right)-(\sqrt{x}+1)(2x)}{(x^2+3)^2}$

$\dfrac{21-4x-7x^2}{(x^2+3)^2}$

You don't need to multiply through.

7

You can use the quotient rule.

$\dfrac{3x^4+30x^3-2x-5}{(x^2+5x)^2}$

$\dfrac{-3x^2+12x+5}{(2-x)^2}$

$\dfrac{-22x}{(3x^2+5)^2}$

$\dfrac{4x^3+12x^2-1}{(x+2)^2}$

There are two pieces of the given function that could cause problems.

The derivative of the function is

The derivative is undefined if either $x<0$ or $x = 0,\pm 1$ (since the square-root is undefined for $x<0$ and the denominator is zero when $x=0,1,-1$. Putting this together — the derivative exists for $x>0, x\neq 1$.

$\sqrt[3]{x}=x^{1/3}$

$\left(\frac{3}{5}{x}^{\frac{-4}{5}}+5{x}^{\frac{-2}{3}}\right)\left(3x^2+8x-5\right)+ \left(3\sqrt[5]{x}+15\sqrt[3]{x}+8\right)\left(6x+8\right)$

Simplify first

$f'(x)=(2x+5)(x^{-1/2}+x^{-2/3})+(x^2+5x+1)\left(\frac{-1}{2}x^{-3/2}-\frac{2}{3}x^{-5/3}\right)$

Differentiate a few times until you get zero, remembering that $a$, $b$, $c$, and $d$ are all constants.

$n=4$

You can re-write this function as a piecewise function, with branches $x \ge 0$ and $x<0$. To figure out the derivatives at $x = 0$, use the definition of a derivative.

(a) In order to make $f(x)$ a little more tractable, let's change the format. Since $|x|=\left\{\begin{array}{rl} x&x \geq 0\\ -x&x<0 \end{array}\right.$, then:

Now, we turn to the definition of the derivative to figure out whether $f'(0)$ exists.

Since $f$ looks different to the left and right of 0, in order to evaluate this limit, we look at the corresponding one-sided limits. Note that when $h$ approaches 0 from the right, $h>0$ so $f(h)=h^2$. By contrast, when $h$ approaches 0 from the left, $h<0$ so $f(h)=-h^2$.

Since both one-sided limits exist and are equal to 0,

and so $f$ is differentiable at $x=0$ and $f'(0)=0$.

(b) From (a), $f'(0)=0$ and

So,

Then, we know the second derivative of $f$ everywhere except at $x=0$:

So, whenever $x \neq 0$, $f''(x)$ exists. To investigate the differentiability of $f'(x)$ when $x=0$, again we turn to the definition of a derivative. If

exists, then $f''(0)$ exists.

Since $f(h)$ behaves differently when $h$ is greater than or less than zero, we look at the one-sided limits.

Since the one-sided limits do not agree,

So, $f''(0)$ does not exist. Now we have a complete picture of $f''(x)$:

Let $m$ be the slope of such a tangent line, and let $P_1$ and $P_2$ be the points where the tangent line is tangent to the two curves, respectively. There are three equations $m$ fulfils: it has the same slope as the curves at the given points, and it is the slope of the line passing through the points $P_1$ and $P_2$.

$y=x-\dfrac{1}{4}$

A line has equation $y=mx+b$, for some constants $m$ and $b$. What has to be true for $y=mb+x$ to be tangent to the first curve at the point $x=\alpha$, and to the second at the point $x=\beta$?

$y=4x-4$ and $y=-2x-1$

Compare this to one of the forms given in the text for the definition of the derivative.