2023.03.23 18:04 *Washingbloddysheets* **Finding the derivative of a function. How do I know when it consist of two functions? And how do I know which formula to use?**

I have to find H'(x), where H(x)=2xsqrt{x}

The solution is 3x (written in my book)

I tried to solve it and got the right solution.

But when I plugged in the problem on “symbolab” and compared the way I solved it to the way symbolab did, it seems like I did it wrong.

Symbolab: So initially it simplified the expression first before finding the derivative. It took the constant 2 out. And found the derivative of x^{{\frac{2}{3}}} - much simpler than what I did

The way I solved it was using the product rule, because I thought 2x could be seen as a function and \sqrt{x} could be seen as a function.

How can I learn from this next time - to know which kind of method to use?

submitted by Washingbloddysheets to learnmath [link] [comments]
The solution is 3x (written in my book)

I tried to solve it and got the right solution.

But when I plugged in the problem on “symbolab” and compared the way I solved it to the way symbolab did, it seems like I did it wrong.

Symbolab: So initially it simplified the expression first before finding the derivative. It took the constant 2 out. And found the derivative of x

The way I solved it was using the product rule, because I thought 2x could be seen as a function and \sqrt{x} could be seen as a function.

How can I learn from this next time - to know which kind of method to use?

2023.02.17 14:36 *your_friend_papu* **Bing Chat confused by a single 'fictional' maths page**

submitted by your_friend_papu to bing [link] [comments] |

2022.12.31 07:15 *WolfeTheMind* **Sqrt((x^3)-4)=x*Sqrt(x-2)**

False input or output from Symbolab. Surely my bad. I thought I was going crazy trying to figure out how it could simplify to that because it's so obvious

Mods delete please. I probably should have just refreshed and retyped first. Still not sure, however, how I messed that up

Very embarrassed. Thanks for the assistance everyone

submitted by WolfeTheMind to askmath [link] [comments]
Mods delete please. I probably should have just refreshed and retyped first. Still not sure, however, how I messed that up

Very embarrassed. Thanks for the assistance everyone

2022.12.02 19:16 *WolfeTheMind* **Do you have to factor first before solving a rational function?**

So I'm trying to find a cross for my horizontal asymptote in my rational function. The asymptote is at 2 and my professor says I can solve using the original equation or the form that's already been factored and canceled out. Symbolab doesn't help as it just simplifies and cancels out factors but when I do the original equation I get a different answer than when I use the factored / cancelled form.

Let me show the example:

2x^{2}-5x+2 / x^{2}-4 = 2 ( the asymptote)

First I multiply 2 by the denominator to get 2x^{2}-5x+2 = 2x^{2}-8

Then I subtract 2x^{2} from both sides and am still left with the variable x in the form of -5x+2 = -8. Which then IS solvable.

However when I factor and cancel (x-2) first I get 2x-1=2x+4, which you can obviously see there is no solution for because the 2x's subtract out

Why am I getting different results for what the prof says is an identical equation (aside from the hole)? What am I getting wrong here

Update: This is the video and time I'm wondering about. He solves for the simplified version but explicitly says that can use either version, which I tried... With different results. That's I guess what I'm trying to figure out

*FINAL UPDATE*: I actually found a youtube comment just now that was asking the same exact thing. Comment:

"1:05:40 P.leonard said that it doesn’t matter whether or not you use the main or simplified version of the function when calculating intersection of the horizontal asymptote, but when i used the main function, i got x=2, meaning that there is an intersection. 2x^2-5x+2=2(x^2-4) 2x^2-5x+2=2x^2-8 -5x+2=-8 -5x=-10, so x=2….. what’s happening here?"

**Top response was very clear, in case anybody is wondering:**

"Great question, but yes, you're not going to get the same answer if its not a cross. In this case you're getting x=2 back showing where (if at all) you're going to have a H.A. cross. But because you know there's a hole at x=2, you know that there wont be a cross at that point. Using the simplified version of the fraction is just a different road to get to the same conclusion. Its just in this case you're getting that "non-sense answer" where two different numbers equal each other. Again great question, because yes, first time trying it you're expecting to get the same answer for both the original function and the simplified version. Two different "roads" that lead to two different "destinations" but the end and final conclusions are consistent "

submitted by WolfeTheMind to askmath [link] [comments]
Let me show the example:

2x

First I multiply 2 by the denominator to get 2x

Then I subtract 2x

However when I factor and cancel (x-2) first I get 2x-1=2x+4, which you can obviously see there is no solution for because the 2x's subtract out

Why am I getting different results for what the prof says is an identical equation (aside from the hole)? What am I getting wrong here

Update: This is the video and time I'm wondering about. He solves for the simplified version but explicitly says that can use either version, which I tried... With different results. That's I guess what I'm trying to figure out

"1:05:40 P.leonard said that it doesn’t matter whether or not you use the main or simplified version of the function when calculating intersection of the horizontal asymptote, but when i used the main function, i got x=2, meaning that there is an intersection. 2x^2-5x+2=2(x^2-4) 2x^2-5x+2=2x^2-8 -5x+2=-8 -5x=-10, so x=2….. what’s happening here?"

"Great question, but yes, you're not going to get the same answer if its not a cross. In this case you're getting x=2 back showing where (if at all) you're going to have a H.A. cross. But because you know there's a hole at x=2, you know that there wont be a cross at that point. Using the simplified version of the fraction is just a different road to get to the same conclusion. Its just in this case you're getting that "non-sense answer" where two different numbers equal each other. Again great question, because yes, first time trying it you're expecting to get the same answer for both the original function and the simplified version. Two different "roads" that lead to two different "destinations" but the end and final conclusions are consistent "

2022.11.04 10:28 *Soggy_Leg_757* **I can't fit my answers into the choices**

Anyways, here are the two problems: 1. What is the first derivative of ln(ln(y)) + ln(y) = ln(x)? A. 2y/(x+y) B. y/(x-y) C. 2y/(x-y) D. y/(x+y)

I tried differentiating it and got y*ln(y)/(x*(1+ln(y))), which is close to choice D. I did it by first simplifying the ln using ln(xy) = ln(x) + ln(y) and then using the product rule. It's an implicit function, therefore, I had to isolate y' after differentiating them, which led me to what I have right now. The problem is, I can't find any way to remove ln to get to fit choice D. I even tried to evaluate each selection using Symbolab by letting it solve the problem initially but got everything wrong. Not a single one of those is correct according to Symbolab. And the answer I got is the same as the one that showed up using Symbolab. Therefore, how?!

submitted by Soggy_Leg_757 to MathHelp [link] [comments]
I tried differentiating it and got y*ln(y)/(x*(1+ln(y))), which is close to choice D. I did it by first simplifying the ln using ln(xy) = ln(x) + ln(y) and then using the product rule. It's an implicit function, therefore, I had to isolate y' after differentiating them, which led me to what I have right now. The problem is, I can't find any way to remove ln to get to fit choice D. I even tried to evaluate each selection using Symbolab by letting it solve the problem initially but got everything wrong. Not a single one of those is correct according to Symbolab. And the answer I got is the same as the one that showed up using Symbolab. Therefore, how?!

- Differentiate y = 2e^x*8^x A. 6.16e^x*8^x B. 2e^x*8^x C. 3e^x*8^x D. 4.16e^x*8^x

2022.09.03 16:36 *BuilderBrilliant* **Can anybody tell me if this is right?**

I am working on simplifying cos3x into cosx and sinx. The z^{n} relation is used. I ended up using a Symbolab calculator for one of the steps, which is in the notebook. The final answer is the last row on the blackboard.

submitted by BuilderBrilliant to askmath [link] [comments]
2022.08.31 16:13 *AnonymColonist* **My maths is just otherworldly**

submitted by AnonymColonist to softwaregore [link] [comments] |

2022.07.26 15:51 *NarcolepticFlarp* **Why Did I Get This Integral Wrong?**

Hello,

I was trying to integrate 1/(x*sqrt(1-x^4)). I used the substitution x^2 = sin(theta). This transformed the integrand into 1/2*csc(theta), which I know how to integrate. After undoing the substitution and simplifying I got ln(x) - 1/2*ln(sqrt(1-x^4) + 1) + c. However, this is not the answer Wolfram Alpha and Symbolab give. What did I do wrong?

Thanks

submitted by NarcolepticFlarp to learnmath [link] [comments]
I was trying to integrate 1/(x*sqrt(1-x^4)). I used the substitution x^2 = sin(theta). This transformed the integrand into 1/2*csc(theta), which I know how to integrate. After undoing the substitution and simplifying I got ln(x) - 1/2*ln(sqrt(1-x^4) + 1) + c. However, this is not the answer Wolfram Alpha and Symbolab give. What did I do wrong?

Thanks

2022.05.14 10:22 *oobeing* **Multivariable limit, stuck at last step of evaluation**

Limit is

lim (x,y) (0,0) x^{2}y^{2}/(x^{2}+y^{4})

I've tried looking if it DNE by going along x and y-axis etc, get 0 on all of them.Changed to polar coordinates and then simplified I get

lim r 0 r^{2}*cos^{2}x*sin^{2}x/(cos^{2}x+r^{2}sin^{4}x)

Then if you plug in you get

0/cos^{2}x

and symbolab says the answer here is 0, because 0/a = 0 when a =/= 0.But how can one know the angle x is not pi/2, making the denominator 0 as well?

Edit: Since limit must be independent upon which path you take for it to exist, meaning it must be independent of the angle x here, wouldn't this mean it doesn't exist since there's one angle/path for which the limit is not defined?

submitted by oobeing to learnmath [link] [comments]
lim (x,y) (0,0) x

I've tried looking if it DNE by going along x and y-axis etc, get 0 on all of them.Changed to polar coordinates and then simplified I get

lim r 0 r

Then if you plug in you get

0/cos

and symbolab says the answer here is 0, because 0/a = 0 when a =/= 0.But how can one know the angle x is not pi/2, making the denominator 0 as well?

Edit: Since limit must be independent upon which path you take for it to exist, meaning it must be independent of the angle x here, wouldn't this mean it doesn't exist since there's one angle/path for which the limit is not defined?

2022.05.14 00:49 *Brown-Banannerz* **A question involving logs**

Simple question

130 = 10log(x/10^-12)

Based on the log rules I'm familiar with, i simplify the equation like so (these are base 1`0 logs)

submitted by Brown-Banannerz to MathHelp [link] [comments]
130 = 10log(x/10^-12)

Based on the log rules I'm familiar with, i simplify the equation like so (these are base 1`0 logs)

- divide both sides by 10: 13 = log(x/10^-12)
- split the log in two components: 13 = log(x) - log(10^-12)
- simplify the second log: 13 = log(x) - -12
- add 12 to both sides: 25 = log(x)
- rearrange the equation based on what a log is: x = 10^25

2022.02.07 23:09 *H4mSandw1ch* **Is there a way to get it to simplify like Symbolab does?**

submitted by H4mSandw1ch to calculators [link] [comments]

2021.11.22 23:04 *awendero* **Is there some kind of trick to solve this algebraic expression?**

I am currently doing preparations for the University entrance exam here in Serbia and I encountered a problem with solving this expression.

I tried all possible ways to simplify it and solve it, but I didn't succeed in that. I tried to evaluate them in Photomath & Symbolab as well, but no luck there either.

I thought it was maybe a printing error, but instead of x I substituted it with -1 (which is the solution) and it really was equal, so I don't know if there is some kind of trick to apply here in order to get a normal solution.

Here is the picture: https://imgur.com/a/jNvMGof

submitted by awendero to cheatatmathhomework [link] [comments]
I tried all possible ways to simplify it and solve it, but I didn't succeed in that. I tried to evaluate them in Photomath & Symbolab as well, but no luck there either.

I thought it was maybe a printing error, but instead of x I substituted it with -1 (which is the solution) and it really was equal, so I don't know if there is some kind of trick to apply here in order to get a normal solution.

Here is the picture: https://imgur.com/a/jNvMGof

2021.10.17 21:15 *Lukaeaeap0* **Integral of ln(x)/x**

I want to calculate the surface between two graphs and I have now simplified this to ∫ ln(x)/x dx. And I need to calculate the surface of the formula between x=1 and x=e.

When searching for an answer I came across this on symbolab%7D%7Bx%7D?or=input) where they substitute u=ln(x), which gives [u^2/2] between x=1 and x=0.

Can someone please explain how this works and how I could apply it myself in other integrals?

submitted by Lukaeaeap0 to learnmath [link] [comments]
When searching for an answer I came across this on symbolab%7D%7Bx%7D?or=input) where they substitute u=ln(x), which gives [u^2/2] between x=1 and x=0.

Can someone please explain how this works and how I could apply it myself in other integrals?

2021.10.15 21:16 *kingtun567* **Modulo Algebra**

I have an assignment that involves me coding using modulo. In one of my functions, I have the equation x = (a * b) mod n. I have to write another function that does the exact same equation, except solved/rewritten for a.

a = ?

How would you go about solving this? I never did modulo algebra in calculus or high school math, so I look at symbolab to see how they solved for a. You can find the solution here. What I found is very strange.

a = (94 mod^23 bx) / (8836 mod^25 b^2); b != 0. I have never seen that notation where the exponent is applied to the modulo or to the left of 'b'. What does it mean? How do they simplify at step 4? How do you write this solution in expanded form so that it can be legally written in code? Thanks for the help.

Edit: I'm also researching into solving for a using modulo identities rather than algebra. So far I've only looked at this: http://pi.math.cornell.edu/~morris/135/mod.pdf, but I cannot determine anything yet.

submitted by kingtun567 to MathHelp [link] [comments]
a = ?

How would you go about solving this? I never did modulo algebra in calculus or high school math, so I look at symbolab to see how they solved for a. You can find the solution here. What I found is very strange.

a = (94 mod^23 bx) / (8836 mod^25 b^2); b != 0. I have never seen that notation where the exponent is applied to the modulo or to the left of 'b'. What does it mean? How do they simplify at step 4? How do you write this solution in expanded form so that it can be legally written in code? Thanks for the help.

Edit: I'm also researching into solving for a using modulo identities rather than algebra. So far I've only looked at this: http://pi.math.cornell.edu/~morris/135/mod.pdf, but I cannot determine anything yet.

2021.09.20 05:43 *dcfan105* **Confused about how to simplify complicated square root expression**

https://drive.google.com/file/d/1yeXaLlOXaaQYPAMS0kMgln2VrCxZX2Sb/view?usp=drivesdk

I'm so confused about how they simplify the first circled expression to get the second circled expression. I've been playing around with this for hours and I just don't get it. I know it has to be right because this is a part of a larger problem involving computing the gradient of a voltage and verifying that it matches the previously derived expression for the electric field. It has to be equal to the second circled expression as that's the expression for the electric field and the problem started with the expression for the voltage, so it's not a mistake on Quizlet's part. But I can't figure out the algebra of going from the first expression to the second. The only thing I can think of is that they somehow factored the denominator of the denominator and canceled out common factors, but if that's what they did I have no idea how.

Whoever wrote this solution apparently thought the step was obvious enough to omit, but it's not at all obvious to me. I even tried putting it into symbolab, hoping it would show me the steps, but it couldn't figure out how to simplify it either, as it claimed it couldn't be simplified.

I'm not even sure what topics to Google to try to find similar problems elsewhere because this is just a very specific type of expression that I don't think has its own name. I don't normally have trouble with algebraic simplification -- heck I'm actually an algebra tutor, but this problem has me utterly stumped and quite frustrated. Can someone please help me understand?

submitted by dcfan105 to askmath [link] [comments]
I'm so confused about how they simplify the first circled expression to get the second circled expression. I've been playing around with this for hours and I just don't get it. I know it has to be right because this is a part of a larger problem involving computing the gradient of a voltage and verifying that it matches the previously derived expression for the electric field. It has to be equal to the second circled expression as that's the expression for the electric field and the problem started with the expression for the voltage, so it's not a mistake on Quizlet's part. But I can't figure out the algebra of going from the first expression to the second. The only thing I can think of is that they somehow factored the denominator of the denominator and canceled out common factors, but if that's what they did I have no idea how.

Whoever wrote this solution apparently thought the step was obvious enough to omit, but it's not at all obvious to me. I even tried putting it into symbolab, hoping it would show me the steps, but it couldn't figure out how to simplify it either, as it claimed it couldn't be simplified.

I'm not even sure what topics to Google to try to find similar problems elsewhere because this is just a very specific type of expression that I don't think has its own name. I don't normally have trouble with algebraic simplification -- heck I'm actually an algebra tutor, but this problem has me utterly stumped and quite frustrated. Can someone please help me understand?

2021.08.09 04:23 *jjrreett* **turning full worksheets into functions/blocks**

I've been using desmos and an engineering ide for a bit now. I just wish it had a bit more functionality. I wish I could turn an entire worksheet into a function or block. Especially so I can use the regression tool as a root finder. Or a million other useful things.

I also wish symbolab was built-in. I often copy and paste equations into symbolab to simplify and solve for among other things.

Finally, a variable editor would be nice. Renaming variables is a hastle.

Desmos is great, wish it was the backbone to a real engineering ide tho (units handling, sig figs handling, error propagation, some spreadsheet things would be nice...

Im interested in Atlas (http://atlasengineering.io/). Can't wait to see how that turns out.

Currently working on building richard nakka's rocket equations into a tool that you tell it a few inputs and it computes grain geometry.

https://www.desmos.com/calculatojisoccndez

submitted by jjrreett to desmos [link] [comments]
I also wish symbolab was built-in. I often copy and paste equations into symbolab to simplify and solve for among other things.

Finally, a variable editor would be nice. Renaming variables is a hastle.

Desmos is great, wish it was the backbone to a real engineering ide tho (units handling, sig figs handling, error propagation, some spreadsheet things would be nice...

Im interested in Atlas (http://atlasengineering.io/). Can't wait to see how that turns out.

Currently working on building richard nakka's rocket equations into a tool that you tell it a few inputs and it computes grain geometry.

https://www.desmos.com/calculatojisoccndez

2021.06.22 19:28 *lilpalp* **Help understanding how (x)(x) equals x^1 here?**

Had to use the Symbolab Simplification calculator to understand where I was making a mistake with my working of this question, I'm confused with the workings shown in one of the steps.

Simplify x(2x^-1/3)^4

= x . 2^4x^-4/3

= 2^4x^1-(4/3)

The calculators method implies that xx^-(4/3) = x^1-(4/3).

Why is it that xx does not equals x^2 here?

submitted by lilpalp to MathHelp [link] [comments]
Simplify x(2x^-1/3)^4

= x . 2^4x^-4/3

= 2^4x^1-(4/3)

The calculators method implies that xx^-(4/3) = x^1-(4/3).

Why is it that xx does not equals x^2 here?

2021.04.25 03:33 *hamzaff44* **I am having a hard time with the following question concerning logarithms.**

How should we simplify a^(log_a (8) + log_a (2))? Thanks for the help.

(solved by finding the answer on symbolab btw)

submitted by hamzaff44 to cheatatmathhomework [link] [comments]
(solved by finding the answer on symbolab btw)

2021.03.15 20:00 *urquhartloch* **How do I solve this equation?**

Ok, so I am studying for the FE next week and I am trying to solve a bunch of FE problems every day. This is from the Math section of the 2015 FE study problems I could find.

Problem: Find the center of the ellipse

((x+y-2)^2)/9+((x-y)^2)/16=1

Solution:

Center: ((x-1)^2)/9+((y-1)^2)/16=1

Therefore the center is at (1,1)

This is all Im given. My problem is that I cant figure out how they got from the initial equation to the solution. I even put it in symbolab and it couldnt output any solution for simplifying it.

submitted by urquhartloch to FE_Exam [link] [comments]
Problem: Find the center of the ellipse

((x+y-2)^2)/9+((x-y)^2)/16=1

Solution:

Center: ((x-1)^2)/9+((y-1)^2)/16=1

Therefore the center is at (1,1)

This is all Im given. My problem is that I cant figure out how they got from the initial equation to the solution. I even put it in symbolab and it couldnt output any solution for simplifying it.

2021.02.28 20:11 *ToeRepresentative627* **Why are these not equivalent expressions? Where did I go wrong in my simplifying?**

I have 2^(2*2^(x-1)), and want to simplify it. I do these steps:

What am I missing? Why are exponent properties not working here like expected?

submitted by ToeRepresentative627 to learnmath [link] [comments]
2^(2*2^x*2^(-1) <---break up the exponent 2^((2*2^x)/(2) <---moved negative exponent to denominator 2^(2^x) <---divided the 2s out. 4^x <---used exponent properties to simplifyHowever... this is not correct. In fact, the correct answer, in its simplest form is 2^(2^x) I have verified this using Symbolab and Desmos which shows indeed that 4^x and 2^(2^x) produce different graphs.

What am I missing? Why are exponent properties not working here like expected?

2021.02.21 19:51 *14thCluelessbird* **Help finding the antiderivative of (5-4x)^3**

I tried using a combination of substitution and the reverse power rule, (x^n+1)/(n+1), but I'm missing something. So I did (u^3+1)(3+1), which simplifies to (u^4)/4. Then I put u back into the function, which got me ((5-4x)^4)/4. However, this is not the right answer. I typed in the equation on symbolab and it looks like you're supposed to multiply the answer I got by -1/4 to get 1/16((5+4x)^4). I'm confused about where the -1/4 came from.

submitted by 14thCluelessbird to askmath [link] [comments]
2021.01.13 14:44 *J-100* **MATLAB Transfer Function Solving Issue**

So I was given a task where I had to get this transfer function diagram, https://gyazo.com/43ca03ef12790167e2cf856a0cb367f7, on MATLAB then optimise it to get a final result. Thing is I solved it on paper previously in order to compare final values and ended up getting this: https://gyazo.com/c6a4508bd02aa40304709160ddc1ac1e/. The next step, I tried following the same process I followed on paper on Matlab by doing the same exact thing; what I did was this:

F1 = tf(2,[0 1 3])

F2 = tf(1,[0 1 8])

F3 = tf(1,[0 1 3])

H1 = tf(4,[0 1 2])

H2 = tf(3,[0 1 12])

H = H1/F3

F = F2 * F3

F = feedback(F,H2,+1)

F = F * F1

F = feedback(F, H)

F = feedback(F, F, +1)

However, to double-check if this was correct, I used a website that was a fraction simplifier to try and solve this same problem, this is what I ended up getting: https://gyazo.com/ec1ae4c2a9d6a34eb0b2b6e057a4b785

Now I'm not sure which is right and which is wrong because I got two totally different answers from both MATLAB and the simplifier website(www.symbolab.com). I would think the website is more accurate and that I must've missed a step or done something incorrectly on Matlab, therefore I would like to know what I did wrong and how it can be fixed in order to get the correct answer. Thank you in advance

submitted by J-100 to matlab [link] [comments]
F1 = tf(2,[0 1 3])

F2 = tf(1,[0 1 8])

F3 = tf(1,[0 1 3])

H1 = tf(4,[0 1 2])

H2 = tf(3,[0 1 12])

H = H1/F3

F = F2 * F3

F = feedback(F,H2,+1)

F = F * F1

F = feedback(F, H)

F = feedback(F, F, +1)

- I essentially followed the same process by firstly moving H1 to the front of F3(becoming H1/F3)
- With the feedback signal now moved, I could freely multiply F2 by F3, storing it in a new variable named F
- F now had the positive feedback of H2 going over it, so I did the feedback thing on Matlab(+1 was used due to the fact that its a positive feedback)
- F is then multiplied by F1 as the path to F from F1 is clear of feedback signals.
- Another feedback, being the original H1/F3 signal is now feedbacked into the main F transfer function. The feedback is negative.
- As you see in the image I linked the first time, there's a single line looping from after the transfer function to the beginning, being a "unity feedback". I somewhat completed this concept by feedbacking F into itself, positively(but i'm not sure it's correct.)

However, to double-check if this was correct, I used a website that was a fraction simplifier to try and solve this same problem, this is what I ended up getting: https://gyazo.com/ec1ae4c2a9d6a34eb0b2b6e057a4b785

Now I'm not sure which is right and which is wrong because I got two totally different answers from both MATLAB and the simplifier website(www.symbolab.com). I would think the website is more accurate and that I must've missed a step or done something incorrectly on Matlab, therefore I would like to know what I did wrong and how it can be fixed in order to get the correct answer. Thank you in advance