- #1

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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

i did this

s

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

sq x *x^2)+4x*sq x+ 3/1

Last edited:

- Thread starter afcwestwarrior
- Start date

- #1

- 457

- 0

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

i did this

s

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

sq x *x^2)+4x*sq x+ 3/1

Last edited:

- #2

- 457

- 0

so what am i suppost to do, i know its suppost to be 3/2 sq x + 2/sqx - 3/2x sq x

- #3

- 457

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any one can u help me

- #4

hage567

Homework Helper

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

- #5

- 17

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Use the Quotient rule.

- #6

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- #7

hage567

Homework Helper

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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.

so what am i suppost to do, i know its suppost to be 3/2 sq x + 2/sqx - 3/2x sq x

- #8

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Ok, Ive never heard of this other way but good luck

- #9

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so x is imaginary?square root of x is a negative number

- #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.

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.

Last edited:

- #11

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

and then differentiate each term .

- #12

HallsofIvy

Science Advisor

Homework Helper

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(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.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

Is that what you meant about square root being "negative"?

- #13

- 457

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yes thats what i meant

- #14

- 457

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

except the problem i posted looks different and is more confusing

- #15

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Learn the quotient rule. The quotient rule is the simplified product rule.

- #16

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- #17

- 123

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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

[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

Last edited:

- #18

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My bad, I didn't see iamlovelyboy's post.

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