Use logarithmic differentiation to find the derivative of y with respect to xy = (10x + 2)^x
Answers
Given: An equation-
[tex]y=(10x+2)^x[/tex]Required: To determine the differentiation of y with respect to x.
Explanation: The differentiation of a logarithmic function is-
[tex]\begin{gathered} y=a^x \\ \frac{dy}{dx}=a^x\ln(a) \end{gathered}[/tex]Taking log both sides on the given equation as-
[tex]\begin{gathered} \ln y=\ln(10x+2)^x \\ =x\ln(10x+2) \end{gathered}[/tex]Now, differentiating with respect to x using product rule as-
[tex]\frac{1}{y}\frac{dy}{dx}=\ln(10x+2)\frac{d}{dx}(x)+x\frac{d}{dx}\ln(10x+2)[/tex]Further simplifying as-
[tex]\frac{dy}{dx}=y[\ln(10x+2)+\frac{10x}{10x+2}][/tex]Substituting the value of y as-
[tex]\frac{dy}{dx}=(10x+2)^x[\ln(10x+2)+\frac{10x}{10x+2}][/tex]Final Answer: Option D is correct.
Related Questions
What is the sum of 1/8+5/16+3/8?
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Firsr we have to make sure that all of the enominators are the same , before we can proceed with the addition. We have to convert the fraction so it willl have the same denominator. We will use the greatest denominator, 16.
[tex]\frac{1}{8}=\frac{\text{?}}{16}[/tex]We can do that by cross multiplication,
[tex]\begin{gathered} \frac{1\cdot16}{8}=\text{?} \\ \questeq2 \\ \end{gathered}[/tex]Thus,
[tex]\frac{1}{8}=\frac{2}{16}[/tex]We, will do the same for 3/8,
[tex]\begin{gathered} \frac{3}{8}=\frac{?}{16} \\ \frac{3\cdot16}{8}=\frac{48}{8}=6 \\ \end{gathered}[/tex]Thus,
[tex]\frac{3}{8}=\frac{6}{16}[/tex]Now that we have the same denominator, we can proceed with additiion.
[tex]\frac{2}{16}+\frac{5}{16}+\frac{6}{16}=\text{?}[/tex]In adding fractions , we just have to add the numerator and copy the common denominator.
[tex]\frac{2}{16}+\frac{5}{16}+\frac{6}{16}=\frac{2+5+6_{}}{16}=\frac{13}{16}[/tex]Answer:
[tex]\frac{13}{16}[/tex]Need help Which expression is equivalent to the given expression?(ab^2)^3/b^OA.a3/bOB.a3boc.a4/bOD.a3
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Given the expression:
[tex]\frac{(ab^2)^3}{b^5}[/tex]We will use the following rules to modify the given expression:
[tex]\begin{gathered} (a^{m)n}=a^{mn} \\ \frac{a^m}{a^n}=a^{m-n} \end{gathered}[/tex]So, the answer will be as follows:
[tex]\frac{(ab^2)^3}{b^5}=\frac{a^3b^6}{b^5}=a^3b^{6-5}=a^3b[/tex]So, the answer will be option ⇒ B. a³b
g(x)=4x^2+3 is the function g even odd or neither? prove it
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Answer:
Even
Step-by-step explanation:
A function is called even if, for all values of x, f(-x) = f(x).
A function is called odd if, for all values of x, f(-x) = -f(x).
If none of these equalities is satisfied, the function is neither even nor odd.
In this question:
g(x) = 4x² + 3
g(-x) = 4(-x)² + 3 = 4x² + 3
Since g(-x) = g(x), the function g is even.
How to find the inverse of a matrix of it exists Question number 15
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2x2 matrix's inverse:
[tex]\begin{gathered} A^{(-1)}=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & {}\end{bmatrix}^{(-1)}=\frac{1}{ad-bc}\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & \end{bmatrix} \\ \\ \\ It\text{ exists only if: } \\ ad-bc\ne0 \end{gathered}[/tex]For the given matrix:
[tex]\begin{gathered} \begin{bmatrix}{6} & {-3} & \\ {-8} & {4} & {}\end{bmatrix} \\ \\ A^{(-1)}=\frac{1}{6\times4-(-3)\times(-8)}\begin{bmatrix}{4} & {3} & {} \\ {8} & {6} & {}\end{bmatrix} \\ \\ A^{(-1)}=\frac{1}{24-24}\begin{bmatrix}{4} & {3} & {} \\ {8} & {6} & {}\end{bmatrix} \\ \\ A^{(-1)}=\frac{1}{0}\begin{bmatrix}{4} & {3} & {} \\ {8} & {6} & {}\end{bmatrix} \\ \\ \end{gathered}[/tex]As the determinat (ad-bc) is 0 the matrix isn't a invertible matrix. The inverse of the given matrix doesn't existFind the coordinates of the other endpoint of the segment, given its midpoint and one endmidpoint (1,-1), endpoint (3,8)
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The midpoint (a, b) = (1, -1)
One endpoint, (x₁, y₁) = (3, 8)
The other endpoint, (x₂, y₂) = ?
Using the formula for midpoint and solving for the missing parameters
[tex]\begin{gathered} a=\frac{x_1+x_2}{2} \\ \\ 1=\frac{3+x_2}{2} \\ \\ 2=3+x_2 \\ \\ x_2=2-3 \\ \\ x_2=-1 \end{gathered}[/tex][tex]\begin{gathered} b=\frac{y_1+y_2}{2} \\ \\ -1=\frac{8+y_2}{2} \\ \\ -2=8+y_2 \\ \\ y_2=-2-8 \\ \\ y_2=-10 \end{gathered}[/tex]The coordinates of the other endpoint = (-1, -10)
Laura is playing a game of chance in which she tosses a dart into a rotating dart board with 8 equal-size slices numbered 1 through 8. The dart lands on a numbered slice at random
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Solution:
Given that;
Laura is playing a game of chance in which she tosses a dart into a rotating dart board with 8 equal-size slices numbered 1 through 8
a) For the expected value;
If Laura tosses the dart once, and she wins $1 if the dart lands in slice 1, $3 if the dart lands in slice 2, $5 if the dart lands in slice 3, $8 if the dart lands in slice 4, and $10 if the dart lands in slice 5 and loses she loses $9 if the dart lands in slices 6, 7, or 8
The expected value will be
[tex]Expected\text{ value}=\frac{1(1)+3(1)+5(1)+8(1)+10(1)+9(3)}{8}=\frac{27-27}{8}=\frac{0}{8}=0[/tex]Hence, the expected value is $0
b) If Laura plays many games, she is expected to neither gain nor lose money.
This is because she has 5 out of 8 chances to win some money and the expected value is $0t
Hence, Laura can expect to break even (neither gain nor lose money)
If PAJB)= PA). state the relationship between events Aand B.
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The condition is:
[tex]P(A\uparrow B)=P(A)[/tex]Then we know that they are overlapping events so is option (D)
plot on a number line . 3| y+4 | >12
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3 |y + 4| > 12
| y + 4 | > 12/3 = 4
| y + 4 | > 4
Two solutions:
y + 4 > 4 ==> y > 0
y + 4 < -4 ==> y < -8
How many 1 --liter bottles of water does it take to fill a 16-liter jug?
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16/1=16
In the equation 3x2 + 6x = 12, the value of c is
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Explanation:
The general quadratic equation is given below as
[tex]ax^2+bx+c=0[/tex]The equation given in the question is given below as
[tex]3x^2+6x=12[/tex]By reaaranging the equation into the standard form, we will have
[tex]3x^2+6x-12=0[/tex]Hence,
The value of c is
[tex]c=-12[/tex]-11. Given points (x, y) and (x2, y2), derive the two-point form of a line. , , , 10. 13. Given that a line is parallel to the x-axis through (x, y), derive the parallel to x-axis form a line.
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11. Given the two points (x1, y1) & (x2, y2) we will have the following line and we derivate it:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex][tex]y-y_1=m(x-x_1)\Rightarrow y=mx-mx_1+y_1[/tex]It's derivative is:
[tex]\frac{\delta y}{\delta x}=m=\frac{y_2-y_1}{x_2-x_1}[/tex]This is since the derivative of constants is 0 and the only variable accompanied m. This is proof that the derivative of a function can be interpreted as the slope of the function at that point.
13. If we have that the line is parallel to the x-axis and passes through the point (x1, y1), we will have that the line is a constant function, so when we derivate no matter the point, it will be equal to 0.
That is:
[tex]y=x_1[/tex][tex]\frac{\delta y}{\delta x}=0[/tex]***Explanation:
point 11:
Since we are given two points (x1, y1) & (x2, y2), we will always have that the slope of the line that passes through those points will always have the form:
Rectangle ABCD is shown on the grid.What is the area of rectangle ABCD in square units?O 3V17 square units6B|(313)06/17 square units2417 square unitsO 34 square units-5.53.234G21.3C (1.-5)Mark this and returnSave and ExitNextSubmit
Answers
The rectangle is:
To find the area of this rectangle, we need to find the distances of the two different sides of the rectangle. We know that the area of a rectangle is given by:
[tex]A_{\text{rectangle}}=w\cdot l_{}[/tex]Then, we need to find the distance between two points for the width, that is, it could be the distance between points C and D (segment CD) or segment AB.
To find the length, we need to find the distance of the segment AD or the distance of the segment BC.
After finding them, we need to multiply the result for w and l, and this product will be the area of the rectangle.
Finding WWe need to apply the formula for the distance between two points:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]A(-1, 4) ---> x1 = -1, y1 = 4
B(3, 3) ---> x2 = 3, y2 = 3
Then, we have:
[tex]d_{AB}=\sqrt[]{(3-(-1))^2+(3-4)^2}=\sqrt[]{(3+1)^2+(-1)^2}=\sqrt[]{(4)^2+(-1)^2}=\sqrt[]{16+1}[/tex]Therefore, this distance, which is w, is equal to:
[tex]d_{AB}=w=\sqrt[]{17}[/tex]Finding LWe can apply the same procedure to find l. We have that:
B(3,3) ---> x1 = 3, y1 = 3
C(1, -5) ---> x2 = 1, y2 = -5
Then, this distance, which is also l is:
[tex]d_{BC}=l=\sqrt[]{(1-3)^2+(-5-3)^2}=\sqrt[]{(-2)^2+(-8)^2}=\sqrt[]{4+64}=\sqrt[]{68}[/tex]Area of the Rectangle ABCDThe area is given by the product of w and l. Then, we have:
[tex]A_{\text{rectangle}}=w\cdot l=\sqrt[]{17}\cdot\sqrt[]{68}[/tex]We know that the factors of 68 are:
[tex]68=2^2\cdot17[/tex]Then, we can rewrite the area as follows:
[tex]A_{\text{rectangle}}=\sqrt[]{17}\cdot\sqrt[]{2^2\cdot17}=\sqrt[]{17}\cdot2\cdot\sqrt[]{17}=2\cdot\sqrt[]{17}\cdot\sqrt[]{17}=2\cdot(17)^{\frac{1}{2}}_{}\cdot(17)^{\frac{1}{2}}[/tex]And, finally, we have:
[tex]A_{\text{rectangle}}=2\cdot(17)^{\frac{1}{2}+\frac{1}{2}}=2\cdot17^1=34\Rightarrow A_{rec\tan gle}=34u^2[/tex]In summary, the area of the rectangle ABCD is equal to 34 square units (last option).
Hello! I need some assistance with this homework question, pleaseQ19
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We start with the parent function:
[tex]y=\sqrt[]{x}[/tex]The transformations are:
• Shift up 6 units
,• Reflect about the x-axis
,• Reflect about the y-axis
The first transformation is equal to add 6 units to y:
[tex]\begin{gathered} y_2=y+6 \\ y_2=\sqrt[]{x}+6 \end{gathered}[/tex]Then, a reflection across the x-axis makes y change sign, so we will have:
[tex]\begin{gathered} y_3=-y_2 \\ y_3=-\sqrt[]{x}-6 \end{gathered}[/tex]Finally, a reflection across the y-axis makes the argument x change sign, so we will have:
[tex]\begin{gathered} y_4=y_3(-x) \\ y_4=-\sqrt[]{-x}-6 \end{gathered}[/tex]This is the final function.
We can graph this function, and the intermediate steps, as:
Answer: y = -√-x - 6
Change the following expression to radical notation: 5x^1/9
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To convert exponential to radical expression and vice versa, we follow the pattern below:
[tex]ab^{\frac{x}{n}}=a\sqrt[n]{b^x}^{}[/tex]In the given term in the question, a = 5, b = x, n = 9, and x = 1. Let's plug these values into the radical form shown above.
The radical form of the given term is:
[tex]5\sqrt[9]{x}[/tex]A room is measured to be 10.0m long, 5.6m wide and 155 m high.What is the volume of the room?What is the mass in kilograms of smoke inside the room if it’s density is 1.506kg/m^3?
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The rule of the volume of the solid shaped a rectangular solid is
[tex]V=L\times W\times H[/tex]L is the length
W is the width
H is the height
Since the dimensions of the room are
10 m long
5.6 m wide
155 m high
Then
L = 10
W = 5.6
H = 155
Substitute them in the rule above
[tex]\begin{gathered} V=10\times5.6\times155 \\ V=8680m^3 \end{gathered}[/tex]The volume of the room is 8680 cubic meters
The rule of the mass is
[tex]m=d\times V[/tex]m is the mass
d is the density
V is the volume
Since the density is 1.506 kg/m^3
Then d = 1.506
Since the volume is 8680 m^3
Then
[tex]\begin{gathered} m=1.506\times8680 \\ m=13072.08\operatorname{kg} \end{gathered}[/tex]The mass of the smoke inside the room is 13072.08 kg
hey, can someone please help me solve this equation? i really need help
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Solution
For this case we can use the following definition:
[tex]\cot N=\frac{\cos N}{\sin N}[/tex]And from the figure given we have:
[tex]\cos N=\frac{12}{37},\sin N=\frac{35}{37}[/tex]And replacing we got:
[tex]\cot N=\frac{12}{35}[/tex]Could you tell me the empty box table what do I put in
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We can see that for x = 0 that implies y = 10.
We can also see that for each x , y decreases 2 units, so the rate of change is -2.
The relation between x and y is a linear function:
y = 10 - 2x
Let's see if it is right.
For x = 0, we get
y = 10 - 2 . 0
y = 10
Ok!
For x = 1
y = 10 - 2 . 1
y = 10 - 2
y = 8
Ok!
For x = 2
y = 10 - 2 . 2
y = 6
Ok!
For x = 15
y = 10 - 2 . 15
y = 10 - 30
y = -20
Answer: start value: 0 ;
rate of change: -2,
relation between x and y: y = 10 -2x
A GPA is usually an example of a a. b. mean bimodal distribution trimodal distribution weighted average C. d. 48
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I find the following data for the concept GPA:
Your GPA is calculated in two steps:
The grade awarded for each course is multiplied by the credit value for each course.
The aggregate score is divided by the total number of credits for all courses completed in the defined period of study.
So, the weighted average reflects the relative contribution made by all the courses you have undertaken based on their credit value.
From the above data, the correct option for the answer is d. weighted average.
For questions 8 - 10, find all the solutions for x.8. 2x 4+14= 229. 8x +19 -54 +3x?10. 7x² + 12 =51- 4x?
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x = 2 or x = -2
Explanation:8) 2x² + 14 = 22
collect like terms by subtracting14 from both sides of the equation:
2x² + 14 - 14 = 22 - 14
2x² + 0 = 8
2x² = 8
DIvide both sides by 2:
2x²/2 = 8/2
x² = 4
Square root both sides of the equation:
[tex]\begin{gathered} \sqrt[]{x^2}\text{ = }\pm\sqrt[]{4} \\ x\text{ =}\pm2 \\ \pm2\text{ = +2 or -2} \\ x\text{ =2 or x = -2} \end{gathered}[/tex]Complete each equation so that it is true for no values of x
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By definition, an equation has no solution when does not exist a value of the variable that makes the equation true.
• In this case, you have this expression on the left side of the first equation:
[tex]3x+6_{}_{}[/tex]Then, knowing part of the right side, you can set up the following:
[tex]3x+6=3(x+1)[/tex]Notice that you need to write a value different from 6, in order to make the equation false, In this case, you cannot complete the missing value with 2, because;
[tex]3\cdot2=6[/tex]Then, having that equation. you can solve it as follows:
[tex]\begin{gathered} 3x+6=3x+3 \\ 3x-3x=3-6 \\ 0=-3\text{ (False)} \end{gathered}[/tex]• Given the second equation that has this left side:
[tex]x-2[/tex]The missing value must be:
If the area of a rectangle is 60 cm and the length is 5cm and the width is (x+4) find x
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Answer:
8
Step-by-step explanation:
A=length*width
60=5*(x+4)
60=5x+20
40=5x subtract 20 from both sides
8=x divide by 5 on both sides
x=8
the temperature at 3pm was 65 degrees it dropped 21 degrees by 7 pm what was the temperature at 7pm
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Given:
it is given that the temperature at 3pm was 65 degrees it dropped 21 degrees by 7 pm.
Solution:
Now, the temperature at 7 pm will be
[tex]65-21=44[/tex]So, temperature at 7 pm is 44 degree.
The decay of a radioactive substance is given byA = 200(1/2) t/10. Which answer is an equivalent equationfor the decay that shows the approximate amount of/decayin 4 years?
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The decay of a radioactive substance is given by
A = 200(1/2) t/10.
The equivalent equation for the decay that shows the approximate amount of decay in 4 years
Hence the correct option that approximately amounts to decay in 4years is
[tex]A=151.57(0.933)^{t-4}[/tex]Option C is the correct answer cause it matches the red image
show the stepsHow do the graphs of y=1/x and y = 5/(x+6) compare?
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• Slightly shifted up
,• Horizontally translated to the left 6 units
1) Consider this:
[tex]y=\frac{1}{x}[/tex]This is the parent function from the familiar of rational functions.
2) Note that on the other hand, the function below:
[tex]y=\frac{5}{x+6}[/tex]This is a transformed function from the first one. Take a look at the graph with both functions plotted:
In Rational functions the greater the numerator the farther from the origin is, shifting up the graph and the addition of 6 in the bottom number translates horizontally to the left.
3) Thus, we can state the answer as:
The graph of y=5/(x+6) is:
• Slightly shifted up
,• Horizontally translated to the left 6 units
Find the measure of angle DAC if point P is the incenter of triangle AEC.
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Given that P is the incenter of the triangle, then angles EAD and DAC are congruent. Therefore, the measure of angle DAC is 33°
The graph of the absolute value function y = -a| x | is reflected over thex-axisy-axisy = xy = -x
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When there is - sign multiplying the absolute value, accompanied by any constant, the graph is reflected over the x-axis. You can confirm this by putting some points in the cartesian plane and plotting them. Therefore, the correct answer is x-axis
What is the solution to the equation below? Round your answer to two decimal places.5x = 55A.x = 1.04B.x = 0.40C.x = 2.49D.x = 2.40
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For this problem, we are given a certain equation and we need to solve it for x.
The expression:
[tex]5^x=55[/tex]We need to apply a logarithm to both sides:
[tex]\log5^x=\log55[/tex]We can now use the properties of the logarithm to isolate the x variable.
[tex]\begin{gathered} x\log5=\log55\\ \\ x=\frac{\log55}{\log5}=2.49 \end{gathered}[/tex]The correct answer is C. x = 2.49
Claim: Most adults erase all of their personal information online if they could. A software firm survey of randomly selected adults showed that % of them would erase all of their personal information online if they could. Make a subjective estimate to decide whether the results are significantly low or significantly high, then state a conclusion about the original claim.
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Claim: Most adults would erase all of their personal information online if they could
Sample size: 625
Yes: 50.3%
The null hypothesis, due to randomness, is p = 0.5. From the results, the claim states that p > 0.5, and the result of the survey is p = 0.503.
The result is close to the null hypothesis, so we can conclude that there is no significant evidence for the claim.
Answer:
The results are not significantly high so there is not sufficient evidence to support the claim.
Determine whether ACDE is similar to AFGHi Answerby i AnswerNoYesAngles not congruentAA similaritySides not proportionalSSS similaritySAS similarity
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Given
Graph of triangles
Procedure
CDE to FGH
[tex]\frac{DE}{GH}=\frac{35}{24}=1.4583[/tex][tex]\frac{DC}{FH}=\frac{28}{20}=1.4[/tex][tex]\frac{CE}{GF}=\frac{21}{16}=1.3125[/tex]Answer NO
SIDES NOT PROPORTIONAL
Solve the equation using the order of operations 4(x-8)+14=-24
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SOLUTION
Using the order of operations we have
[tex]\text{PEMDAS}[/tex]where
[tex]\begin{gathered} P=\text{parenthesis} \\ E=\text{Exponential } \\ M=\text{ Multiplication } \\ D=\text{Division } \\ A=\text{Addition } \\ S=\text{Subtraction } \end{gathered}[/tex]Note: Multiplication and Division operate at the same level but we consider the operation that appears first from the left hand sides of the equation given the same as Addition and subtraction.
Given the equation.
[tex]4(x-8)+14=-24[/tex]Expand the parenthesis
[tex]4x-32+14=-24[/tex]Add the like terms
[tex]4x-18=-24[/tex]Add 18 to both sides of the equation
[tex]\begin{gathered} 4x-18+18=-24+18 \\ 4x=-6 \end{gathered}[/tex]Divide both sides by 4
[tex]\begin{gathered} \frac{4x}{4}=\frac{-6}{4} \\ \\ x=\frac{-3}{2} \end{gathered}[/tex]Therefore
x= -3/2