need help

[jas14]

i tried but still can't get it

what are the derivatives of the following:
1. square root of x^4+9x^2
i got (4x^3 + 18x) / (2 sqrt x^4 + 9x^2)

2. cosx^2 / 3x (x is squared, not cosine)
i got -2xsinx^2 / 3

3. 4^(3x-1)^2 or 4^(9x^2-6x+1)
i got (18x-6)ln(4)

 

[soroban]

Hello, jas14!

A few errors . . .

\(\displaystyle 1)\;y\;=\;\sqrt{x^4\,+\,9x^2\)

Your answer is correct . . . it can be reduced.

We have: .\(\displaystyle y\;=\;(x^4\,+\,9x^2)^{\frac{1}{2}}\)

Then: .\(\displaystyle y'\;=\;\frac{1}{2}\cdot(x^4\,+\,9x^2)^{-\frac{1}{2}}\cdot(4x^3\,+\,18x)\;=\;\frac{2x^3\,+\,9x}{\sqrt{x^4\,+\,9x^2}}\)

\(\displaystyle 2)\;y\;=\;\frac{\cos(x^2)}{3x}\)

Quotient Rule: .\(\displaystyle y'\;=\;\frac{3x\cdot[-\sin(x^2)]\cdot2x\,-\,\cos(x^2)\cdot3}{(3x)^2}\;=\;\frac{-6x^2\cdot\sin(x^2)\,-\,3\cdot\cos(x^2)}{9x^2}\)

. . . \(\displaystyle =\;\frac{3[-2x^2\cdot\sin(x^2)\,-\,\cos(x^2)]}{9x^2}\;=\;\frac{-2x^2\cdot\sin(x^2)\,-\,\cos(x^2)}{3x^2}\)

\(\displaystyle 3)\;y\;=\;4^{(3x-1)^2}\;=\;4^{(9x^2-6x+1)}\)

Formula: . If \(\displaystyle y\:=\:b^u\), then: .\(\displaystyle y'\:=\:b^u\cdot u'\cdot\ln(b)\)

Your answer: .\(\displaystyle y\;=\;4^{(9x^2-6x+1)}\cdot(18x\,-\,6)\cdot\ln(4)\)

 

[jas14]

???

umm... someone told me that the ( cosx^2 / 3x ) problem is incorrect because there aren't supposed to be parenthesis around the x^2, so my answer would be different

 

[soroban]

Re: ???

There aren't supposed to be parenthesis around the x^2.

Great! . . . Let's try one more time . . .
[And how come you're not using the Quotient Formula?}

We have: .\(\displaystyle y\;=\;\frac{\cos^2(x)}{3x}\)

Then: .\(\displaystyle y'\;=\;\frac{(3x)\cdot2\cdot\cos(x)\cdot[-\sin(x)]\,-\,[\cos^2(x)]\cdot3}{(3x)^2}\;=\;\frac{-6x\cdot\sin(x)\cdot\cos(x)\,-\,3\cdot\cos^2(x)}{9x^2}\)

. . . . . \(\displaystyle y'\;=\;\frac{-3\cdot\cos(x)\cdot[2\cdot\sin(x)\,+\,\cos(x)]}{9x^2}\;=\;\frac{-\cos(x)\cdot[2\cdot\sin(x)\,+\,\cos(x)]}{3x^2}\)

*

 

[jas14]

?

why did u change it from cosx^2 to cos^2x?

 

[Gene]

You can't use too many ()'s to make it clear what you mean. If Sorobans interpretation is correct the prefered way would be
cos(x^2)/(3x)
Strictly obeying the rules what you typed (modified by your explanation, which put implied ()'s around x^2) would equal
x*cos(x^2)/3
Any thing inder a / should be in ()'s
Any thing above a / should be in ()'s if there is a + or - in it
All finctions should be in ()'s.
You might do a web-search on PEMDAS for the complete rules for reading what you type for correctness.
If you use a TI calculator and click "cos" it puts "cos(" automatically. I disagree with the critic.

 

[jas14]

confused

so is cos^2(x) the same as cosx^2?
i still dont know what the answer to the problem is. what is the derivative of
cosx^2
divided by
3x

 

[pka]

“so is cos<SUP>2</SUP>(x) the same as cos(x<SUP>2</SUP>)?
If you do not know the answer to that question, you have no business in a calculus course
You ought take my advise: Rethink your being in calculus!

 

[jas14]

i no that these are 2 different thingsso soroban's answer is wrong because he used cos^2 but its actually cosx^2

 

[pka]

“i no that these are 2 different things”
Do you mean
“I know that these are 2 different thing…”
If you know that, then why ask?
I think you just want the correct answer so that you can get the credit for some else’s work!

 

[happy]

Again, pka is right on the money! You tell 'em, pka! You made this person look like a fool twice today, and I think you should be commended for it. :wink: