Ap Calculus Bc Mixed Review Pdf Substitution Natural Log Logarithmic Differentiation

April 18, 2022 Post a Comment

3. Derivatives

3.9 Derivatives of Exponential and Logarithmic Functions

Learning Objectives

Notice the derivative of exponential functions.
Notice the derivative of logarithmic functions.
Utilise logarithmic differentiation to determine the derivative of a function.

So far, we take learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As nosotros discussed in Introduction to Functions and Graphs, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions.

Derivative of the Exponential Function

Merely as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to brand certain basic assumptions. The proofs that these assumptions concord are across the telescopic of this class.

Showtime of all, nosotros begin with the assumption that the part $B(x)=b^x, \, b>0$ , is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined—showtime with the definition of $b^n$ , where $n$ is a positive integer—every bit the product of $b$ multiplied past itself $n$ times. Later, we defined $b^0=1, \, b^{−n}=\frac{1}{b^n}$ for a positive integer $n$ , and $b^{s/t}=(\sqrt[t]{b})^s$ for positive integers $s$ and $t$ . These definitions leave open the question of the value of $b^r$ where $r$ is an arbitrary real number. By bold the continuity of $B(x)=b^x, \, b>0$ , we may interpret $b^r$ as $\underset{x\to r}{\lim}b^x$ where the values of $x$ as we have the limit are rational. For example, we may view ${4}^{\pi}$ every bit the number satisfying

$\begin{array}{l}4^3<4^{\pi}<4^4, \, 4^{3.i}<4^{\pi}<4^{iii.ii}, \, 4^{three.14}<4^{\pi}<4^{3.xv},\\ 4^{3.141}<4^{\pi}<4^{iii.142}, \, 4^{3.1415}<iv^{\pi}<4^{3.1416}, \, \cdots \end{array}$

As we see in the following table, $4^{\pi}\approx 77.88$ .

Approximating a Value of $4^{\pi}$
$x$	$4^x$	$x$	$4^x$
$4^3$	64	$4^{3.141593}$	77.8802710486
$4^{3.1}$	73.5166947198	$4^{3.1416}$	77.8810268071
$4^{3.14}$	77.7084726013	$4^{3.142}$	77.9242251944
$4^{3.141}$	77.8162741237	$4^{3.15}$	78.7932424541
$4^{3.1415}$	77.8702309526	$4^{3.2}$	84.4485062895
$4^{3.14159}$	77.8799471543	$4^4$	256

We also assume that for $B(x)=b^x, \, b>0$ , the value $B^{\prime}(0)$ of the derivative exists. In this section, we show that by making this one boosted assumption, it is possible to prove that the function $B(x)$ is differentiable everywhere.

Nosotros make one final assumption: that there is a unique value of $b>0$ for which $B^{\prime}(0)=1$ . We ascertain $e$ to be this unique value, as we did in Introduction to Functions and Graphs. (Figure) provides graphs of the functions $y=2^x, \, y=3^x, \, y=2.7^x$ , and $y=2.8^x$ . A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of $e$ lies somewhere between 2.vii and 2.8. The office $E(x)=e^x$ is called the natural exponential office. Its changed, $L(x)=\log_e x=\ln x$ is called the natural logarithmic function.

For a better estimate of $e$ , we may construct a table of estimates of $B^{\prime}(0)$ for functions of the form $B(x)=b^x$ . Earlier doing this, recall that

$B^{\prime}(0)=\underset{x\to 0}{\lim}\frac{b^x-b^0}{x-0}=\underset{x\to 0}{\lim}\frac{b^x-1}{x} \approx \frac{b^x-1}{x}$

for values of $x$ very close to nix. For our estimates, we choose $x=0.00001$ and $x=-0.00001$ to obtain the judge

$\frac{b^{-0.00001}-1}{-0.00001}<B^{\prime number}(0)<\frac{b^{0.00001}-1}{0.00001}$ .

Meet the following tabular array.

Estimating a Value of $e$ $b$ $\frac{b^{-0.00001}-ane}{-0.00001}<B^{\prime}(0)<\frac{b^{0.00001}-1}{0.00001}$ $b$ $\frac{b^{-0.00001}-1}{-0.00001}<B^{\prime}(0)<\frac{b^{0.00001}-1}{0.00001}$ 2 $0.693145<B^{\prime}(0)<0.69315$ 2.7183 $1.000002<B^{\prime}(0)<1.000012$ 2.seven $0.993247<B^{\prime}(0)<0.993257$ 2.719 $1.000259<B^{\prime}(0)<1.000269$ 2.71 $0.996944<B^{\prime}(0)<0.996954$ 2.72 $ane.000627<B^{\prime number}(0)<1.000637$ 2.718 $0.999891<B^{\prime number}(0)<0.999901$ 2.viii $1.029614<B^{\prime}(0)<1.029625$ 2.7182 $0.999965<B^{\prime}(0)<0.999975$ iii $1.098606<B^{\prime}(0)<1.098618$

The testify from the table suggests that $2.7182<e<2.7183$ .

The graph of $E(x)=e^x$ together with the line $y=x+1$ are shown in (Figure). This line is tangent to the graph of $E(x)=e^x$ at $x=0$ .

At present that we accept laid out our basic assumptions, we begin our investigation by exploring the derivative of $B(x)=b^x, \, b>0$ . Retrieve that we have assumed that $B^{\prime}(0)$ exists. By applying the limit definition to the derivative we conclude that

$B^{\prime}(0)=\underset{h\to 0}{\lim}\frac{b^{0+h}-b^0}{h}=\underset{h\to 0}{\lim}\frac{b^h-1}{h}$ .

Turning to $B^{\prime}(x)$ , we obtain the following.

$\begin{array}{lllll} B^{\prime}(x) & =\underset{h\to 0}{\lim}\frac{b^{x+h}-b^x}{h} & & & \text{Apply the limit definition of the derivative.} \\ & =\underset{h\to 0}{\lim}\frac{b^xb^h-b^x}{h} & & & \text{Note that} \, b^{x+h}=b^x b^h. \\ & =\underset{h\to 0}{\lim}\frac{b^x(b^h-1)}{h} & & & \text{Factor out} \, b^x. \\ & =b^x\underset{h\to 0}{\lim}\frac{b^h-1}{h} & & & \text{Apply a property of limits.} \\ & =b^x B^{\prime}(0) & & & \text{Use} \, B^{\prime}(0)=\underset{h\to 0}{\lim}\frac{b^{0+h}-b^0}{h}=\underset{h\to 0}{\lim}\frac{b^h-1}{h}. \end{array}$

Nosotros come across that on the footing of the assumption that $B(x)=b^x$ is differentiable at $0, \, B(x)$ is not only differentiable everywhere, but its derivative is

$B^{\prime}(x)=b^x B^{\prime}(0)$ .

For $E(x)=e^x, \, E^{\prime}(0)=1$ . Thus, we have $E^{\prime}(x)=e^x$ . (The value of $B^{\prime}(0)$ for an arbitrary role of the class $B(x)=b^x, \, b>0$ , will be derived later.)

Derivative of the Natural Exponential Office

Let $E(x)=e^x$ be the natural exponential function. Then

$E^{\prime}(x)=e^x$ .

In general,

$\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\prime}(x)$ .

Derivative of an Exponential Function

Find the derivative of $f(x)=e^{\tan (2x)}$ .

Solution

Using the derivative formula and the chain rule,

$\begin{array}{ll} f^{\prime}(x) & =e^{\tan (2x)}\frac{d}{dx}(\tan (2x)) \\ & = e^{\tan (2x)} \sec^2 (2x) \cdot 2. \end{array}$

Combining Differentiation Rules

Observe the derivative of $y=\frac{e^{x^2}}{x}$ .

Solution

Use the derivative of the natural exponential part, the quotient dominion, and the chain rule.

$\begin{array}{lllll} y^{\prime} & =\large \frac{(e^{x^2} \cdot 2x) \cdot x - 1 \cdot e^{x^2}}{x^2} & & & \text{Apply the quotient rule.} \\ & = \large \frac{e^{x^2}(2x^2-1)}{x^2} & & & \text{Simplify.} \end{array}$

Find the derivative of $h(x)=xe^{2x}$ .

Solution

$h^{\prime}(x)=e^{2x}+2xe^{2x}$

Applying the Natural Exponential Function

A colony of mosquitoes has an initial population of 1000. After $t$ days, the population is given by $A(t)=1000e^{0.3t}$ . Show that the ratio of the rate of modify of the population, $A^{\prime}(t)$ , to the population size, $A(t)$ is constant.

Solution

First observe $A^{\prime}(t)$ . By using the chain dominion, we take $A^{\prime}(t)=300e^{0.3t}$ . Thus, the ratio of the charge per unit of alter of the population to the population size is given by

$\large \frac{A^{\prime}(t)}{A(t)} \normalsize = \large \frac{300e^{0.3t}}{1000e^{0.3t}}=0.3$ .

The ratio of the charge per unit of modify of the population to the population size is the constant 0.3.

Derivative of the Logarithmic Function

Now that we take the derivative of the natural exponential function, we can apply implicit differentiation to find the derivative of its inverse, the natural logarithmic function.

The Derivative of the Natural Logarithmic Office

If $x>0$ and $y=\ln x$ , then

$\frac{dy}{dx}=\frac{1}{x}$ .

More generally, let $g(x)$ be a differentiable function. For all values of $x$ for which $g^{\prime}(x)>0$ , the derivative of $h(x)=\ln(g(x))$ is given past

$h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x)$ .

Proof

If $x>0$ and $y=\ln x$ , then $e^y=x$ . Differentiating both sides of this equation results in the equation

$e^y\frac{dy}{dx}=1$ .

Solving for $\frac{dy}{dx}$ yields

$\frac{dy}{dx}=\frac{1}{e^y}$ .

Finally, we substitute $x=e^y$ to obtain

$\frac{dy}{dx}=\frac{1}{x}$ .

We may also derive this result by applying the inverse part theorem, as follows. Since $y=g(x)=\ln x$ is the inverse of $f(x)=e^x$ , by applying the inverse function theorem we have

$\frac{dy}{dx}=\frac{1}{f^{\prime}(g(x))}=\frac{1}{e^{\ln x}}=\frac{1}{x}$ .

Using this outcome and applying the concatenation rule to $h(x)=\ln(g(x))$ yields

$h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x). _\blacksquare$

The graph of $y=\ln x$ and its derivative $\frac{dy}{dx}=\frac{1}{x}$ are shown in (Figure).

Taking a Derivative of a Natural Logarithm

Observe the derivative of $f(x)=\ln(x^3+3x-4)$ .

Solution

Use (Figure) directly.

$\begin{array}{lllll} f^{\prime}(x) & =\frac{1}{x^3+3x-4} \cdot (3x^2+3) & & & \text{Use} \, g(x)=x^3+3x-4 \, \text{in} \, h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x). \\ & =\frac{3x^2+3}{x^3+3x-4} & & & \text{Rewrite.} \end{array}$

Using Backdrop of Logarithms in a Derivative

Detect the derivative of $f(x)=\ln(\frac{x^2 \sin x}{2x+1})$ .

Solution

At outset glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the trouble much simpler.

$\begin{array}{lllll} f(x) & = \ln(\frac{x^2 \sin x}{2x+1})=2\ln x+\ln(\sin x)-\ln(2x+1) & & & \text{Apply properties of logarithms.} \\ f^{\prime}(x) & = \frac{2}{x} + \frac{\cos x}{\sin x} -\frac{2}{2x+1} & & & \text{Apply sum rule and} \, h^{\prime}(x)=\frac{1}{g(x)} g^{\prime}(x). \\ & = \frac{2}{x} + \cot x - \frac{2}{2x+1} & & & \text{Simplify using the quotient identity for cotangent.} \end{array}$

Differentiate: $f(x)=\ln (3x+2)^5$ .

Solution

$f^{\prime}(x)=\frac{15}{3x+2}$

At present that we can differentiate the natural logarithmic function, nosotros tin can use this result to observe the derivatives of $y=\log_b x$ and $y=b^x$ for $b>0, \, b\ne 1$ .

Derivatives of General Exponential and Logarithmic Functions

Let $b>0, \, b\ne 1$ , and let $g(x)$ exist a differentiable office.

If $y=\log_b x$ , so
$\frac{dy}{dx}=\frac{1}{x \ln b}$ .

More by and large, if $h(x)=\log_b (g(x))$ , then for all values of $x$ for which $g(x)>0$ ,

$h^{\prime}(x)=\frac{g^{\prime}(x)}{g(x) \ln b}$ .
If $y=b^x$ , then
$\frac{dy}{dx}=b^x \ln b$ .

More by and large, if $h(x)=b^{g(x)}$ , then

$h^{\prime}(x)=b^{g(x)} g^{\prime}(x) \ln b$ .

Proof

If $y=\log_b x$ , then $b^y=x$ . Information technology follows that $\ln(b^y)=\ln x$ . Thus $y \ln b = \ln x$ . Solving for $y$ , we take $y=\frac{\ln x}{\ln b}$ . Differentiating and keeping in mind that $\ln b$ is a abiding, we meet that

$\frac{dy}{dx}=\frac{1}{x \ln b}$ .

The derivative in (Figure) now follows from the chain dominion.

If $y=b^x$ , so $\ln y=x \ln b$ . Using implicit differentiation, once more keeping in mind that $\ln b$ is constant, it follows that $\frac{1}{y}\frac{dy}{dx}=\text{ln}b.$ Solving for $\frac{dy}{dx}$ and substituting $y=b^x$ , we come across that

$\frac{dy}{dx}=y \ln b=b^x \ln b$ .

The more than general derivative ((Effigy)) follows from the chain rule. $_\blacksquare$

Applying Derivative Formulas

Detect the derivative of $h(x)=\large \frac{3^x}{3^x+2}$ .

Solution

Use the quotient rule and (Figure).

$\begin{array}{lllll} h^{\prime}(x) & = \large \frac{3^x \ln 3(3^x+2)-3^x \ln 3(3^x)}{(3^x+2)^2} & & & \text{Apply the quotient rule.} \\ & = \large \frac{2 \cdot 3^x \ln 3}{(3^x+2)^2} & & & \text{Simplify.} \end{array}$

Finding the Slope of a Tangent Line

Find the slope of the line tangent to the graph of $y=\log_2 (3x+1)$ at $x=1$ .

Solution

To notice the slope, we must evaluate $\frac{dy}{dx}$ at $x=1$ . Using (Figure), we encounter that

$\frac{dy}{dx}=\frac{3}{\ln 2(3x+1)}$ .

By evaluating the derivative at $x=1$ , we see that the tangent line has slope

$\frac{dy}{dx}|_{x=1} =\frac{3}{4 \ln 2}=\frac{3}{\ln 16}$ .

Find the slope for the line tangent to $y=3^x$ at $x=2$ .

Solution

$9 \ln (3)$

Logarithmic Differentiation

At this point, we can take derivatives of functions of the form $y=(g(x))^n$ for certain values of $n$ , too every bit functions of the form $y=b^{g(x)}$ , where $b>0$ and $b\ne 1$ . Unfortunately, we notwithstanding do not know the derivatives of functions such as $y=x^x$ or $y=x^{\pi}$ . These functions crave a technique called logarithmic differentiation, which allows us to differentiate any function of the grade $h(x)=g(x)^{f(x)}$ . It can also be used to catechumen a very complex differentiation trouble into a simpler i, such as finding the derivative of $y=\frac{x\sqrt{2x+1}}{e^x \sin^3 x}$ . Nosotros outline this technique in the following trouble-solving strategy.

Using Logarithmic Differentiation

Find the derivative of $y=(2x^4+1)^{\tan x}$ .

Solution

Use logarithmic differentiation to observe this derivative.

$\begin{array}{lllll} \ln y & = \ln (2x^4+1)^{\tan x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = \tan x \ln (2x^4+1) & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = \sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x & & & \begin{array}{l}\text{Step 3. Differentiate both sides. Use the} \\ \text{product rule on the right.} \end{array} \\ \frac{dy}{dx} & =y \cdot (\sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x) & & & \text{Step 4. Multiply by} \, y \, \text{on both sides.} \\ \frac{dy}{dx} & = (2x^4+1)^{\tan x}(\sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x) & & & \text{Step 5. Substitute} \, y=(2x^4+1)^{\tan x}.\end{array}$

Using Logarithmic Differentiation

Observe the derivative of $y=\large \frac{x\sqrt{2x+1}}{e^x \sin^3 x}$ .

Solution

This problem really makes use of the properties of logarithms and the differentiation rules given in this affiliate.

$\begin{array}{lllll} \ln y & = \ln \large \frac{x\sqrt{2x+1}}{e^x \sin^3 x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = \ln x+\frac{1}{2} \ln (2x+1)-x \ln e-3 \ln \sin x & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = \frac{1}{x}+\frac{1}{2x+1}-1-3\big(\frac{\cos x}{\sin x}\big) & & & \text{Step 3. Differentiate both sides.} \\ \frac{dy}{dx} & = y (\frac{1}{x}+\frac{1}{2x+1}-1-3 \cot x) & & & \text{Step 4. Multiply by} \, y \, \text{on both sides and simplify.} \\ \frac{dy}{dx} & = \large \frac{x\sqrt{2x+1}}{e^x \sin^3 x} \normalsize (\frac{1}{x}+\frac{1}{2x+1}-1-3 \cot x) & & & \text{Step 5. Substitute} \, y=\large \frac{x\sqrt{2x+1}}{e^x \sin^3 x}. \end{array}$

Extending the Power Rule

Detect the derivative of $y=x^r$ where $r$ is an arbitrary existent number.

Solution

The procedure is the aforementioned as in (Effigy), though with fewer complications.

$\begin{array}{lllll} \ln y & = \ln x^r & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = r \ln x & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = r \frac{1}{x} & & & \text{Step 3. Differentiate both sides.} \\ \frac{dy}{dx} & = y \frac{r}{x} & & & \text{Step 4. Multiply by} \, y \, \text{on both sides.} \\ \frac{dy}{dx} & = x^r \frac{r}{x} & & & \text{Step 5. Substitute} \, y=x^r. \\ \frac{dy}{dx} & = rx^{r-1} & & & \text{Simplify.} \end{array}$

Use logarithmic differentiation to find the derivative of $y=x^x$ .

Solution

$\frac{dy}{dx}=x^x(1+\ln x)$

Find the derivative of $y=(\tan x)^{\pi}$ .

Solution

$y^{\prime}=\pi (\tan x)^{\pi -1} \sec^2 x$

Key Concepts

Fundamental Equations

For the following exercises, find $f^{\prime}(x)$ for each function.

1. $f(x)=x^2 e^x$

Solution

$f^{\prime}(x) = 2xe^x+x^2 e^x$

two. $f(x)=\frac{e^{−x}}{x}$

three. $f(x)=e^{x^3 \ln x}$

Solution

$f^{\prime}(x) = e^{x^3 \ln x}(3x^2 \ln x+x^2)$

iv. $f(x)=\sqrt{e^{2x}+2x}$

five. $f(x)=\large \frac{e^x-e^{−x}}{e^x+e^{−x}}$

Solution

$f^{\prime}(x) = \large \frac{4}{(e^x+e^{−x})^2}$

6. $f(x)=\frac{10^x}{\ln 10}$

7. $f(x)=2^{4x}+4x^2$

Solution

$f^{\prime}(x) = 2^{4x+2} \cdot \ln 2+8x$

viii. $f(x)=3^{\sin 3x}$

ix. $f(x)=x^{\pi} \cdot \pi^x$

Solution

$f^{\prime}(x) = \pi x^{\pi -1} \cdot \pi^x + x^{\pi} \cdot \pi^x \ln \pi$

ten. $f(x)=\ln(4x^3+x)$

11. $f(x)=\ln \sqrt{5x-7}$

Solution

$f^{\prime}(x) = \frac{5}{2(5x-7)}$

12. $f(x)=x^2 \ln 9x$

13. $f(x)=\log(\sec x)$

Solution

$f^{\prime}(x) = \frac{\tan x}{\ln 10}$

14. $f(x)=\log_7 (6x^4+3)^5$

xv. $f(x)=2^x \cdot \log_3 7^{x^2-4}$

Solution

$f^{\prime}(x) = 2^x \cdot \ln 2 \cdot \log_3 7^{x^2-4} + 2^x \cdot \frac{2x \ln 7}{\ln 3}$

For the post-obit exercises, utilize logarithmic differentiation to find $\frac{dy}{dx}$ .

16. $y=x^{\sqrt{x}}$

17. $y=(\sin 2x)^{4x}$

Solution

$\frac{dy}{dx} = (\sin 2x)^{4x} [4 \cdot \ln(\sin 2x) + 8x \cdot \cot 2x]$

18. $y=(\ln x)^{\ln x}$

19. $y=x^{\log_2 x}$

Solution

$\frac{dy}{dx} = x^{\log_2 x} \cdot \frac{2 \ln x}{x \ln 2}$

twenty. $y=(x^2-1)^{\ln x}$

21. $y=x^{\cot x}$

Solution

$\frac{dy}{dx} = x^{\cot x} \cdot [−\csc^2 x \cdot \ln x+\frac{\cot x}{x}]$

22. $y= \large \frac{x+11}{\sqrt[3]{x^2-4}}$

23. $y=x^{-1/2}(x^2+3)^{2/3}(3x-4)^4$

Solution

$\frac{dy}{dx} = x^{-1/2}(x^2+3)^{2/3}(3x-4)^4 \cdot [\frac{-1}{2x}+\frac{4x}{3(x^2+3)}+\frac{12}{3x-4}]$

24. [T] Observe an equation of the tangent line to the graph of $f(x)=4xe^{x^2-1}$ at the point where

$x=-1$ . Graph both the office and the tangent line.

25. [T] Find the equation of the line that is normal to the graph of $f(x)=x \cdot 5^x$ at the point where $x=1$ . Graph both the function and the normal line.

Solution

The function starts at (−3, 0), decreases slightly and then increases through the origin and increases to (1.25, 10). There is a straight line marked T(x) with slope −1/(5 + 5 ln 5) and y intercept 5 + 1/(5 + 5 ln 5).
$y=\frac{-1}{5+5 \ln 5}x+(5+\frac{1}{5+5 \ln 5})$

29. [T] The population of Toledo, Ohio, in 2000 was approximately 500,000. Presume the population is increasing at a rate of 5% per year.

Write the exponential function that relates the total population as a role of $t$ .
Use a. to determine the rate at which the population is increasing in $t$ years.
Use b. to determine the rate at which the population is increasing in x years.

Solution

a. $P=500,000(1.05)^t$ individuals
b. $P^{\prime}(t)=24395 \cdot (1.05)^t$ individuals per year
c. 39,737 individuals per year

Solution

a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York Urban center. At the beginning of 1963 there were approximately 723 cases of the disease in New York City.
b. At the commencement of 1960 the number of cases of the disease was decreasing at rate of -4.611 1000 per year; at the start of 1963, the number of cases of the disease was decreasing at a rate of -0.2808 thousand per year.

For the following exercises, use the population of New York Metropolis from 1790 to 1860, given in the post-obit table.

Years since 1790	Population
0	33,131
10	60,515
20	96,373
30	123,706
40	202,300
50	312,710
sixty	515,547
70	813,669

33. [T] Using a figurer program or a calculator, fit a growth curve to the information of the form $p=ab^t$ .

Solution

$p=35741(1.045)^t$

34. [T] Using the exponential best fit for the data, write a table containing the derivatives evaluated at each year.

35. [T] Using the exponential best fit for the data, write a table containing the second derivatives evaluated at each year.

Solution

Years since 1790	$P''$
0	69.25
10	107.5
20	167.0
30	259.4
40	402.eight
50	625.5
lx	971.four
70	1508.5

36. [T] Using the tables of commencement and second derivatives and the best fit, answer the post-obit questions:

Volition the model be accurate in predicting the future population of New York City? Why or why not?
Estimate the population in 2010. Was the prediction correct from a.?

Glossary

logarithmic differentiation: is a technique that allows u.s. to differentiate a office by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly

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Source: https://opentextbc.ca/calculusv1openstax/chapter/derivatives-of-exponential-and-logarithmic-functions/

Ap Calculus Bc Mixed Review Pdf Substitution Natural Log Logarithmic Differentiation

3.9 Derivatives of Exponential and Logarithmic Functions

Learning Objectives

Derivative of the Exponential Function

Derivative of the Natural Exponential Office

Derivative of an Exponential Function

Solution

Combining Differentiation Rules

Solution

Solution

Applying the Natural Exponential Function

Solution

Derivative of the Logarithmic Function

The Derivative of the Natural Logarithmic Office

Proof

Taking a Derivative of a Natural Logarithm

Solution

Using Backdrop of Logarithms in a Derivative

Solution

Solution

Derivatives of General Exponential and Logarithmic Functions

Proof

Applying Derivative Formulas

Solution

Finding the Slope of a Tangent Line

Solution

Solution

Logarithmic Differentiation

Using Logarithmic Differentiation

Solution

Using Logarithmic Differentiation

Solution

Extending the Power Rule

Solution

Solution

Solution

Key Concepts

Fundamental Equations

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Glossary

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