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D/dx 6x^(3/2)tan(x) = 3x^(1/2)(2xsec^2(x)+3tan(x))? ?

  • Thread starter zketrouble
  • Start date
d/dx 6x^(3/2)tan(x) = 3x^(1/2)(2xsec^2(x)+3tan(x))??!?!??!?!!!?

Hi all,

I've been officially bored to death with my College Algebra class so I just started teaching myself Calculus about two months ago. I found a list of problems using the product rule http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html#PROBLEM 4". Problem 4 is as follows:

Differentiate 6x^(3/2)tan(x)

Doesn't look too hard, I start by remembering that (fg)' = f'g +fg'.

d/dx f(x) = d/dx[6x^(3/2)]tan(x) + 6x^(3/2)d/dx[tan(x)]
d/dx f(x) = 9x^(1/2)tan(x) + 6x^(3/2)sec^2(x)

So I clicked the link on that site to view the detailed solution, which says the answer is 3x^(1/2)(2xsec^2(x)+3tan(x)). I hate to make the assumption that it is incorrect being a calculus noob and all, but it appears that 3x^(1/2)(2xsec^2(x)+3tan(x)) is incorrectly factored, because 3x^(1/2) times 3tan(x) does not equal 6x^(3/4)tan(x). The websites solution says that my answer is the step before they arrive at this solution which I assume to be false, am I missing something here?


Thanks!
 
Last edited by a moderator:
Re: d/dx 6x^(3/2)tan(x) = 3x^(1/2)(2xsec^2(x)+3tan(x))??!?!??!?!!!?

3x^(1/2)(2xsec^2(x)+3tan(x)) is correctly factored. Work it out: 3^(1/2)*3=9^(1/2) and 2x*3x^(1/2)=6x^(3/2) I'm not sure where you're getting 6x^(3/4)tan(x). But other than that you did everything correctly.
 

VietDao29

Homework Helper
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Re: d/dx 6x^(3/2)tan(x) = 3x^(1/2)(2xsec^2(x)+3tan(x))??!?!??!?!!!?

Hi all,

I've been officially bored to death with my College Algebra class so I just started teaching myself Calculus about two months ago. I found a list of problems using the product rule http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/productruledirectory/ProductRule.html#PROBLEM 4". Problem 4 is as follows:

Differentiate 6x^(3/2)tan(x)

Doesn't look too hard, I start by remembering that (fg)' = f'g +fg'.

d/dx f(x) = d/dx[6x^(3/2)]tan(x) + 6x^(3/2)d/dx[tan(x)]
d/dx f(x) = 9x^(1/2)tan(x) + 6x^(3/2)sec^2(x)

So I clicked the link on that site to view the detailed solution, which says the answer is 3x^(1/2)(2xsec^2(x)+3tan(x)). I hate to make the assumption that it is incorrect being a calculus noob and all, but it appears that 3x^(1/2)(2xsec^2(x)+3tan(x)) is incorrectly factored, because 3x^(1/2) times 3tan(x) does not equal 6x^(3/4)tan(x). The websites solution says that my answer is the step before they arrive at this solution which I assume to be false, am I missing something here?


Thanks!
[tex]3x^{\frac{1}{2}} \left( 2x\sec ^ 2 (x) + 3 \tan x \right) = 3x^{\frac{1}{2}} 2x\sec ^ 2 (x) + 3 \times 3x^{\frac{1}{2}} \tan x[/tex]
[tex]= 6x^{\frac{1}{2} + 1}\sec ^ 2 (x) + 9x^{\frac{1}{2}} \tan x = 6x^{\frac{3}{2}}\sec ^ 2 (x) + 9x^{\frac{1}{2}} \tan x[/tex]

Looks good to me. :) And it looks exactly as what you've got, too.
 
Last edited by a moderator:
Re: d/dx 6x^(3/2)tan(x) = 3x^(1/2)(2xsec^2(x)+3tan(x))??!?!??!?!!!?

3x^(1/2)(2xsec^2(x)+3tan(x)) is correctly factored. Work it out: 3^(1/2)*3=9^(1/2) and 2x*3x^(1/2)=6x^(3/2) I'm not sure where you're getting 6x^(3/4)tan(x). But other than that you did everything correctly.
Ok silly mistake on my part...just like nearly all of my mistakes. Thanks.
 

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