Question

CAN SOMEONE WALK ME THROUGH THE STEPS OF SIMPLYING THIS PROBLEM.

X^2[(4X-3)^-3/4 (4)] +(4X-3) ^1/4 (2X)

Answer 1

\[x^2[(4x-3)^{\frac{-3}{4}}(4)]+(4x-3)^\frac{1}{4}(2x) ?\]

Answer 2

YES

Answer 3

I guess you want to write as one term and without negative exponents.

Answer 4

So I would start by factoring the (4x-3)^(-3/4) out

Answer 5

YES POSITIVE EXPONENTS ONLY

Answer 6

\[(4x-3)^\frac{-3}{4}[ x^2(4)+(4x-3)^\frac{?}{?}(2x)]\]
what number n satisfies this equation:
-3/4+n=1/4

Answer 7

1

Answer 8

right so that ? I have there is 1

Answer 9

\[(4x-3)^\frac{-3}{4}[x^2(4)+(4x-3)^1(2x)] \\ \text{ but we don't really need \to write 1} \\ \text{ since } v^1 =v \\ (4x-3)^\frac{-3}{4}[x^2(4)+(4x-3)(2x)]\]

Answer 10

now just try to see if you can combine any like terms in the [ ]
and then put that one thing with the negative exponent underneath therefore making the exponent positive

Answer 11

\[ (4x-3)^{-3/4} we get -3x+9/4 \]

Answer 12

?

Answer 13

I don't know what you did with the exponent to get you that

Answer 14

\[(4x-3)^\frac{-3}{4}[x^2(4)+(4x-3)(2x)] \\ \frac{[x^2(4)+(4x-3)(2x)]}{(4x-3)^\frac{3}{4}} \\ \frac{x^2(4)+(4x-3)(2x)}{(4x-3)^\frac{3}{4}}\]
ok I will let you do the adding of like terms if any on top

Answer 15

I multiply the exponnets by whats inside. Iam having trouble getting the answer that my book is stating which is. \[\frac{ 3x(3x-2) }{ (4x-3)^{3/4} }\]

Answer 16

\[5^2 \text{ doesn't meant } 5 \cdot 2 \]

Answer 17

also based on the answer in your book our top will not be equal to the top they have
so you might want to check if I understand what you wrote above

Answer 18

is this the exact problem:
\[x^2[(4x-3)^{\frac{-3}{4}}(4)]+(4x-3)^\frac{1}{4}(2x) ?\]

Answer 19

but it anyways I think your main problem is dealing with rewriting of the \[u^\frac{-3}{4}+u^\frac{1}{4} \\ \text{ I factored this by taking out the } u^\frac{-3}{4} \\ u^\frac{-3}{4}(1+u^\frac{4}{4}) \\ u^\frac{-3}{4}(1+u^1) \\ u^\frac{-3}{4}(1+u) \\ \text{ now recall } x^{-a} \text{ can be written as } \frac{1}{x^a} \\ \frac{1+u}{u^\frac{3}{4}}\]

Answer 20

\[x^{2}[\frac{ 1 }{ 4 }(4x-3)^{-3/4}(4)] +(4x-3)(2x)\]

Answer 21

still factor out the (4x-3)^(-3/4) part

Answer 22

I think you left out the 1/4 for that one exponent
anyways I think I know what the initial problem was find the evaluate of
\[(x^2(4x-3)^\frac{1}{4})'\]
\[=x^2 \frac{1}{4}(4x-3)^{\frac{1}{4}-\frac{4}{4}}(4)+2x(4x-3)^\frac{1}{4} \\ =x^2 \frac{4}{4}(4x-3)^\frac{-3}{4}+2x(4x-3)^\frac{1}{4} \\ =x^2(4x-3)^\frac{-3}{4}+2x(4x-3)^\frac{1}{4} \\ \\ \text{ we will come back \to this } \\ \text{ say we have } \\ f(x) \cdot u^\frac{-3}{4}+g(x) \cdot u^\frac{1}{4} \\ \text{ we can factor out the } u^\frac{-3}{4} \\ u^\frac{-3}{4}(f(x)+g(x) \cdot u^\frac{4}{4}) \text{ this is \because } \frac{-3}{4}+\frac{4}{4}=\frac{1}{4} \\ \text{ now that one factor has a negative exponent } \\ \text{ so recall we can write } x^{-a} \text{ as } \frac{1}{x^a} \\ \frac{1}{u^\frac{3}{4}}(f(x)+g(x) \cdot u^1 ) \text{ since } \frac{4}{4}=1 \\ \text{ so we have } \frac{f(x)+g(x) \cdot u }{u^\frac{3}{4}}\]

Answer 23

Please let me know if you can't follow something above.

Answer 24

yea the beggining throws me off 1/4- 4/4?

Answer 25

do you know the power rule?

Answer 26

\[(x^n)'=nx^{n-1}\]

Answer 27

I took away 1

Answer 28

1=4/4

Answer 29

I wrote 1 as 4/4 so I could have a common denominator

Answer 30

got it. Iam looking up some videos that will help undertand better, Thank you soo much for your help.

Answer 31

If you don't like how I wrote it as one term, I could show you another way.

Answer 32

\[x^2(4x-3)^\frac{-3}{4}+2x(4x-3)^\frac{1}{4}\]
so like you have up til this part I believe...
\[\frac{x^2}{(4x-3)^\frac{3}{4}}+2x(4x-3)^\frac{1}{4} \\ \text{ now we want \to combine the fractions } \\ \frac{x^2}{(4x-3)^\frac{3}{4}}+\frac{2x(4x-3)^\frac{1}{4} \cdot (4x-3)^\frac{3}{4}}{(4x-3)^\frac{3}{4}} \\ \frac{x^2+2x(4x-3)^\frac{1}{4} (4x-3)^\frac{3}{4}}{(4x-3)^\frac{3}{4}}\]

Answer 33

now you know 1/4+3/4=4/4=1

Answer 34

\[\frac{x^2+2x(4x-3)^{\frac{1}{4}+\frac{3}{4}}}{(4x-3)^\frac{3}{4}} \\ \frac{x^2+2x(4x-3)}{(4x-3)^\frac{3}{4}}\]

Answer 35

now distribute on top and combine like terms