derivaive equations

[Guest]

k I need help simplifying what I did OR I am just doing horribly on these equations as I keep getting the wrong answers, help please ^_^.

b) g(x)= x^1/2(x^3-x)
g'(x)= 1/2x^-(1/2)(x^3-x)+x^1/2(3x^2-1)
g'(x)=x^3-x/2(x square rooted)+3x^5/2-xsquare rooted.
I don't know what to do next, the ans is xsquarerooted(7x^2-3)/2

d) y=[3-(1/x)]/x
y=[3-x^-1]/x
y'(x)= [1x^-2(x)-(3-x^-1)(1)]/ x^2
y'(x)= [x^-1-3+x^-1]/x^2
y'(x)= [(2x^-1)-3]/x^2
y'(x)=2-3/x^3
y'(x)= -1/x^3 <--WRONG ANS, its suppose to be [-3x+2]/x^3 so could you tell me where I went wrong?

i)y= (x^2+5x+4)/(x+4)
y'(x)= [2x+5(x+4)-(x^2+5x+4)(1)]/ (x+4)^2
y'(x)= [2x+5x+20-x^2-5x-4]/ (x+4)^2
y'(x)= [-x^2+2x+16/] (x+4)^2
The ans is 1, x can not equal 4, Why is this the answer? that's so weird!

Thanks for the help in advance, and just a warning, I will probably have more questions tonight, because there's a test tomorrow =0

 

[galactus]

k I need help simplifying what I did OR I am just doing horribly on these equations as I keep getting the wrong answers, help please ^_^.

b) g(x)= x^1/2(x^3-x)
g'(x)= 1/2x^-(1/2)(x^3-x)+x^1/2(3x^2-1)
g'(x)=x^3-x/2(x square rooted)+3x^5/2-xsquare rooted.
I don't know what to do next, the ans is xsquarerooted(7x^2-3)/2

d) y=[3-(1/x)]/x
y=[3-x^-1]/x
y'(x)= [1x^-2(x)-(3-x^-1)(1)]/ x^2
y'(x)= [x^-1-3+x^-1]/x^2
y'(x)= [(2x^-1)-3]/x^2
y'(x)=2-3/x^3
y'(x)= -1/x^3 <--WRONG ANS, its suppose to be [-3x+2]/x^3 so could you tell me where I went wrong?
I'll step through this one. See if you spot the problem. OK?.

\(\displaystyle \L\\\frac{3-\frac{1}{x}}{x}=\frac{3x-1}{x^{2}}\)

Quotient rule:

\(\displaystyle \L\\\frac{x^{2}(3)-(3x-1)(2x)}{x^{4}}\)

\(\displaystyle \L\\\frac{2x-3x^{2}}{x^{4}}\)

\(\displaystyle \L\\\frac{-3}{x^{2}}+\frac{2}{x^{3}}\)

Multiply top and bottom of left side by x to get common denominator:

\(\displaystyle \L\\\frac{(x)}{(x)}\frac{-3}{x^{2}}+\frac{2}{x^{3}}\)

There ya' go:

\(\displaystyle \L\\\frac{2-3x}{x^{3}}\)

i)y= (x^2+5x+4)/(x+4)
y'(x)= [2x+5(x+4)-(x^2+5x+4)(1)]/ (x+4)^2
y'(x)= [2x+5x+20-x^2-5x-4]/ (x+4)^2
y'(x)= [-x^2+2x+16/] (x+4)^2
The ans is 1, x can not equal 4, Why is this the answer? that's so weird!

Thanks for the help in advance, and just a warning, I will probably have more questions tonight, because there's a test tomorrow =0

 

[Guest]

oh okay thanks, that's easier when you simplify it at the beginning, now can anyone help with the next two questions =p
and how do I factor this, : -6x(1-x^2)^2(6+2x)^-3-[6(1-x^2)^3(6+2x)^-4
I know you can take a '6' out and a (1-x^2)^2 out, but how about (6+2x)^?

 

[stapel]

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

How did you go from "[(2/x) - 3] / x²" to "2 - (3/x³)"?

i)y= (x^2+5x+4)/(x+4)
y'(x)= [2x+5(x+4)-(x^2+5x+4)(1)]/ (x+4)^2

Why did you multiply the denominator, x + 4, only on the second term of the derivative of the numerator? (This might be one of those times to use more parentheses, because I think that's where you're having problems.)

Eliz.

 

[soroban]

Hello, bittersweet!

Negative and fractional exponents are always tricky to simplify . . .

\(\displaystyle b)\;g(x)\:=\:x^{\frac{1}{2}}(x^3\,-\,x)\)

\(\displaystyle g'(x)\:=\:\frac{1}{2}x^{-\frac{1}{2}}(x^3\,-\,x)\,+\,x^{\frac{1}{2}}(3x^2\,-\,1)\;\;\) . . . correct!

The answer is: \(\displaystyle \,\frac{x^{\frac{1}{2}}(7x^2\,-\,3)}{2}\)

Multiply top and bottom by \(\displaystyle 2x^{\frac{1}{2}}:\)

\(\displaystyle \L\;\;g'(x)\;=\;\frac{2x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}\,\cdot\,\frac{\frac{1}{2}x^{-\frac{1}{2}}(x^3\,-\,x)\,+\,x^{\frac{1}{2}}(3x^2 - 1)}{1} \;= \;\frac{(x^3\,-\,x)\,+\,2x(3x^2\,-\,1)}{2x^{\frac{1}{2}}}\)

\(\displaystyle \L\;\;g'(x)\;=\;\frac{x^3\,-\,x\,+\,6x^3\,-\,2x}{2x^{\frac{1}{2}}} \;= \;\frac{7x^3\,-\,3x}{2x^{\frac{1}{2}}\)

Factor: \(\displaystyle \L\,g'(x)\;=\;\frac{x(7x^2\,-\,3)}{2x^{\frac{1}{2}}}\;=\;\frac{x^{\frac{1}{2}}(7x^2\,-\,3)}{2}\)

\(\displaystyle \L d)\;y\:=\:\frac{3\,-\,\frac{1}{x}}{x}\)

\(\displaystyle \L y\:=\:\frac{3\,-\,x^{^{-1}}}{x}\)

\(\displaystyle \L y'\:=\:\frac{(x)(+1x^{^{-2}})\,-\,(3\,-\,x^{^{-1}})(1)}{x^{^2}}\)

\(\displaystyle \L y'\:=\:\frac{x^{^{-1}}\,-\,3\,+\,x^{^{-1}}}{x^{^2}}\)

\(\displaystyle \L y'\:=\:\frac{2x^{^{-1}}\,-\,3}{x^{^2}}\;\;\) . . . You're one step away

Answer: \(\displaystyle \L\,\frac{-3x\,+\,2}{x^{^3}}\)

You have: \(\displaystyle \L\,\frac{2x^{^{-1}}\,-\,3}{x^{^2}}\)

Multiply top and bottom by \(\displaystyle x:\L\;\;\frac{x}{x}\,\cdot\,\frac{2x^{^{-1}}\,-\,3}{x^{^2}} \;= \;\frac{2\,-\,3x}{x^{^3}}\)

\(\displaystyle i)\;y\:=\:\frac{x^2\,+\,5x\,+\,4}{x\,+\,4}\)

The answer is: \(\displaystyle \,1,\;x\,\neq\,-4\)

This is a sneaky one . . . but it's something to watch for.

Note that the original function can be factored and simplified.

\(\displaystyle \L\;\;y\;=\;\frac{x^{^2}\,+\,5x\,+\,4}{x\,+\,4}\;=\;\frac{(x\,+\,1)(x\,+\,4)}{x\,+\,4}\)

If we promise that \(\displaystyle \,x\,\neq\,-4\), we can cancel.
\(\displaystyle \;\;\)(But the original function says that \(\displaystyle x\,\neq\,-4\), doesn't it?)

So the function becomes: \(\displaystyle \,y\:=\:x\,+\,1\)

Therefore: \(\displaystyle \,y'\:=\:1\;\;(x\,\neq\,-4)\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Your method should have worked . . . but you made some errors.

Quotient Rule: \(\displaystyle \L\:y'\;=\;\frac{(x\,+\,4)(2x\,+\,5)\,-\,(x^2\,+5x\,+\,4)\cdot 1}{(x\,+\,4)^2}\)

\(\displaystyle \L\;\;y'\;=\;\frac{2x^2\,+\,13x\,+\,20\,-\,x^2\,-\,5x\,-\,4}{(x\,+\,4)^2} \;= \;\frac{x^2\,+\,8x\,+\,16}{(x\,+\,4)^2}\)

\(\displaystyle \L\;\;y'\;=\;\frac{(x\,+\,4)^2}{(x\,+\,4)^2}\;=\;1\;\;\) . . . see?