simplify the answer

[al-horia]

hi,
I need help in how to simplify my answer

f(x) = 3y+1 / [FONT=MathJax_Main]√ 1-4y
f(x) = 3y + 1 / (1-4y)^1/2

f '(x) = ( 1-4y)^1/2 (3) - (3y+1) [ 1/2 (1-4y)^-1/2 -4 ] / (( 1-4)^1/2)²

f'(x) = 3(1-4y)^1/2 + 2(3y+1) (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main](( 1-4)[/FONT]^1/2[FONT=MathJax_Main])[/FONT]²

I think we can simplify the answer by this way

[FONT=MathJax_Main]f'(x) = 3 + 2(3y+1) (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2
[FONT=MathJax_Main]f'(x) = 3 + 6y+2 (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2
[FONT=MathJax_Main]f'(x) = 6y+5 (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2[FONT=MathJax_Main]

can we simplify it more ?
[/FONT]

 

[Deleted member 4993]

hi,
I need help in how to simplify my answer

f(x) = 3y+1 / [FONT=MathJax_Main]√ 1-4y
f(x) = 3y + 1 / (1-4y)^1/2

f '(x) = ( 1-4y)^1/2 (3) - (3y+1) [ 1/2 (1-4y)^-1/2 -4 ] / (( 1-4)^1/2)²

f'(x) = 3(1-4y)^1/2 + 2(3y+1) (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main](( 1-4)[/FONT]^1/2[FONT=MathJax_Main])[/FONT]²

I think we can simplify the answer by this way

[FONT=MathJax_Main]f'(x) = 3 + 2(3y+1) (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2
[FONT=MathJax_Main]f'(x) = 3 + 6y+2 (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2
[FONT=MathJax_Main]f'(x) = 6y+5 (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2[FONT=MathJax_Main]

can we simplify it more ?
[/FONT]

I cannot figure out your work - you need to use grouping symbols properly.

If you wanted to write:

\(\displaystyle \dfrac{3y \ + \ 1}{\sqrt{1\ \ - \ 4y}}\)

in ASCII text, you should have written that as:

(3y + 1)/√(1 - 4y)

Now please use parentheses and fix your post.

 

[srmichael]

hi,
I need help in how to simplify my answer

f(x) = 3y+1 / [FONT=MathJax_Main]√ 1-4y
f(x) = 3y + 1 / (1-4y)^1/2

f '(x) = ( 1-4y)^1/2 (3) - (3y+1) [ 1/2 (1-4y)^-1/2 -4 ] / (( 1-4)^1/2f'(x) = 3(1-4y)^1/2 + 2(3y+1) (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main](( 1-4)[/FONT]^1/2[FONT=MathJax_Main])[/FONT]²

I think we can simplify the answer by this way

[FONT=MathJax_Main]f'(x) = 3 + 2(3y+1) (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2
[FONT=MathJax_Main]f'(x) = 3 + 6y+2 (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2
[FONT=MathJax_Main]f'(x) = 6y+5 (1-4y)^-1/2 / [/FONT][FONT=MathJax_Main]( 1-4)[/FONT]^1/2[FONT=MathJax_Main]

can we simplify it more ?
[/FONT]

What happened to the y in the denominator in your 3rd step? I agree with SK, it's a little hard to follow, but I think this is what you are striving for:


\(\displaystyle f(x)=\frac{3y+1}{\sqrt{1-4y}}\)

\(\displaystyle f'(x)=\frac{(3)\sqrt{1-4y}-(3y+1)[\frac{1}{2}(1-4y)^{\frac{-1}{2}}(-4)]}{1-4y}\)

\(\displaystyle =\frac{(3)\sqrt{1-4y}+\frac{2(3y+1)}{\sqrt{1-4y}}}{1-4y}\)

\(\displaystyle =\frac{(3)(1-4y)+2(3y+1)}{(1-4y)^{\frac{3}{2}}}\)

\(\displaystyle =\frac{3-12y+6y+2}{(1-4y)^{\frac{3}{2}}}\)

\(\displaystyle =\frac{5-6y}{(1-4y)^{\frac{3}{2}}}\)

 

[al-horia]

I cannot figure out your work - you need to use grouping symbols properly.

If you wanted to write:

\(\displaystyle \dfrac{3y \ + \ 1}{\sqrt{1\ \ - \ 4y}}\)

in ASCII text, you should have written that as:

(3y + 1)/√(1 - 4y)

Now please use parentheses and fix your post.

from where can I find ASCII text ?

 

[mmm4444bot]

from where can I find ASCII text ?

You might have confused ASCII with LaTex.

"ASCII" is acronym -- it mainly refers to the basic characters available when you type on computer.

"LaTex" is the name for system of mathematical-formatting codes that converts ASCII characters to pretty-math display.

This is ASCII → (3y + 1)/√(1 - 4y)

This is LaTex → \(\displaystyle \dfrac{3y \ + \ 1}{\sqrt{1\ \ - \ 4y}}\)

For information about using LaTex to display pretty math, please google keywords: latex tutorials


When you post math using ASCII characters only, you MUST include the grouping symbols. :cool:

 

[al-horia]

You might have confused ASCII with LaTex.

"ASCII" is acronym -- it mainly refers to the basic characters available when you type on computer.

"LaTex" is the name for system of mathematical-formatting codes that converts ASCII characters to pretty-math display.

This is ASCII → (3y + 1)/√(1 - 4y)

This is LaTex → \(\displaystyle \dfrac{3y \ + \ 1}{\sqrt{1\ \ - \ 4y}}\)

For information about using LaTex to display pretty math, please google keywords: latex tutorials


When you post math using ASCII characters only, you MUST include the grouping symbols. :cool:

thank you dear

 

[al-horia]

What happened to the y in the denominator in your 3rd step? I agree with SK, it's a little hard to follow, but I think this is what you are striving for:


\(\displaystyle f(x)=\frac{3y+1}{\sqrt{1-4y}}\)

\(\displaystyle f'(x)=\frac{(3)\sqrt{1-4y}-(3y+1)[\frac{1}{2}(1-4y)^{\frac{-1}{2}}(-4)]}{1-4y}\)

\(\displaystyle =\frac{(3)\sqrt{1-4y}+\frac{2(3y+1)}{\sqrt{1-4y}}}{1-4y}\)

\(\displaystyle =\frac{(3)(1-4y)+2(3y+1)}{(1-4y)^{\frac{3}{2}}}\) how this happened ? how did you drop the square root ? and how the power become 3/2 ?

\(\displaystyle =\frac{3-12y+6y+2}{(1-4y)^{\frac{3}{2}}}\)

\(\displaystyle =\frac{5-6y}{(1-4y)^{\frac{3}{2}}}\)

thank you very very much for your help

 

[mmm4444bot]

thank you dear

You're welcome, dear.

To view the actual LaTex codes, right-click on the pretty math, and select Show Math As ► TeX Commands

When posting LaTex coding, each line needs to be placed inside [ֺtex] and [ֺ/tex] tags.

:cool:

 

[mmm4444bot]

\(\displaystyle \frac{(3)(1-4y)+2(3y+1)}{(1-4y)^{\frac{3}{2}}}\)

how this happened ? how did you drop the square root ? and how the power become 3/2 ?

Soroban multiplied both the top and bottom by the radical. :cool:

 

[JeffM]

\(\displaystyle =\frac{(3)\sqrt{1-4y}+\frac{2(3y+1)}{\sqrt{1-4y}}}{1-4y}\)

\(\displaystyle =\frac{(3)(1-4y)+2(3y+1)}{(1-4y)^{\frac{3}{2}}}\)

how this happened ? how did you drop the square root ? and how the power become 3/2 ?

\(\displaystyle \frac{(3)\sqrt{1- 4y}+\frac{2(3y+1)}{\sqrt{1- 4y}}}{1- 4y} = \dfrac{1}{(1 - 4y)^1} * \left(3\sqrt{1 - 4y} + \dfrac{2(3y + 1)}{\sqrt{1 - 4y}}\right) =\)

\(\displaystyle \dfrac{1}{(1 - 4y)^{(2/2)}} * \dfrac{3\left(\sqrt{1 - 4y}\right)^2 + 2(3y + 1)}{(1 - 4t)^{(1/2)}} = \dfrac{3(1 - 4y) + 6y + 2}{(1 - 4y)^{(3/2)}}\)