Differentiate this function

  • #1
y=x^2+4x+3/square root x

i did this

s
x^2+4x+3) square root x/ sq x * sq x

sq x *x^2)+4x*sq x+ 3/1
 
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Answers and Replies

  • #2
i made all the sq x equal 1/2x^-1/2
so what am i suppost to do, i know its suppost to be 3/2 sq x + 2/sqx - 3/2x sq x
 
  • #4
hage567
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I can't tell what the function is. Is the square root of x just with the 3, or is it the denominator to all of it?? Try to use brackets to make things clearer.

Is this it? [tex]y=(x^2+4x+3)/\sqrt{x}[/tex]
 
  • #5
Use the Quotient rule.
 
  • #6
thats it, on this one u dont use the quotient rule, i could, but there's a way since square root of x is a negative number
 
  • #7
hage567
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i made all the sq x equal 1/2x^-1/2
so what am i suppost to do, i know its suppost to be 3/2 sq x + 2/sqx - 3/2x sq x
OK, I THINK i see what you're getting at here, but I don't understand why you are multiplying sq x * sq x. If you break up each term to be over sq x, you can just then simplify the x each term, and then take the derivative. This way you don't have to do the quotient rule, you are just taking the derivative of each term separately. That is the only way I can see getting the answer you say it is supposed to be.
 
  • #8
Ok, Ive never heard of this other way but good luck
 
  • #9
11
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thats it, on this one u dont use the quotient rule, i could, but there's a way since square root of x is a negative number
so x is imaginary?:confused:
 
  • #10
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I am in calculus 1 as well so correct me if I'm wrong. I would approach this problem a different way. First off quotient rule is too messy for my liking, Id rather use the product rule. To do this:

Step 1: Bring the denominator up so you can use the product rule:

Since:
sqrt(x) = x^1/2 and 1/(x^1/2) = x^-1/2 you can move sqrt(x) to the top (moving the x^1/2 to the numerator makes the exponent negative)

The original function (x^2+4x+3)/sqrt(x) now equals (x^2+4x+3)*(x^-1/2)
Step 2: Now, using the product rule solve:
(x^2+4x+3)*(-1/2)(x^-3/2)+(x^-1/2)*(2x+4)
simplify:
-1(x^2+4x+3)/2(x^3/2)+(2x+4)/(x^1/2)
To get even denominators multiply the right side by 2x
(-x^2-4x-3)/2(x^3/2)+(4x^2+8x)/2(x^3/2)
simplify, giving the derivative:
(3x^2+4x-3)/(2x^3/2)

*edit* if you write this out you will understand it ALOT better, I havnt mastered how to do the coding yet.
 
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  • #11
use this

use (a+b+c)/d = a/d + b/d + c/d

and then differentiate each term .
 
  • #12
HallsofIvy
Science Advisor
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y=x^2+4x+3/square root x

i did this

s
x^2+4x+3) square root x/ sq x * sq x

sq x *x^2)+4x*sq x+ 3/1
(x^4+ 4x+ 3)/sqrt(x)= (x^4+ 4x+ 3)/x^(1/2)= (x^4+ 4x+ 3)x^(-1/2) and you can use the product rule.

Is that what you meant about square root being "negative"?
 
  • #14
actually in this equation u use the power rule,

take 1/x^2= -1x^-2= -1/x^2

except the problem i posted looks different and is more confusing
 
  • #15
370
0
Learn the quotient rule. The quotient rule is the simplified product rule.
 
  • #16
If you want a simpler way, you could divide the equation (as in y= (x^2)/(x^.5) +4x/(x^.5) +(3x^0)/(x^.5), and simplify the exponents (subtract them). Then just use the general power rule to find the derivative.
 
  • #17
123
0
iamlovelyboy's method avoids the quotient and the product rule

[tex]y=(x^2+4x+3)/\sqrt{x}

=(x^2 + 4x + 3)x^{-1/2}

= x^{3/2} + 4x^{1/2} + 3x^{-1/2}[/tex]

now differentiate each term
 
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  • #18
My bad, I didn't see iamlovelyboy's post.
 

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