differentiate 3000√(x^2 +9) + 7000√(125 - 20x + x^2)

[hndalama]

3000√(x² +9) + 7000√(125 - 20x + x²)

the answer i get is 3000x / √(x²+9) + 7000(10 - x) / √(125 - 20x + x²)

the answer that is given is 3000x / √(x²+9) - 7000(10 - x) / √(125 - 20x + x²)

My question is where does the minus sign come from?

 

[Deleted member 4993]

3000√(x² +9) + 7000√(125 - 20x + x²)

the answer i get is 3000x / √(x²+9) + 7000(10 - x) / √(125 - 20x + x²)

the answer that is given is 3000x / √(x²+9) - 7000(10 - x) / √(125 - 20x + x²)

My question is where does the minus sign come from?

d/dx [125 - 20x + x²] = -2*(10 - x)

Book's answer is correct!!

 

[Steven G]

3000√(x² +9) + 7000√(125 - 20x + x²)

the answer i get is 3000x / √(x²+9) + 7000(10 - x) / √(125 - 20x + x²)

the answer that is given is 3000x / √(x²+9) - 7000(10 - x) / √(125 - 20x + x²)

My question is where does the minus sign come from?

Where did you get 10-x from? I get 2x-20 which become 10-x when you divide by 2. Not sure why the -sign was factored out by your text or teacher. In math you want to use the least number of neg signs.

 

[hndalama]

d/dx [125 - 20x + x²] = -2*(10 - x)

Book's answer is correct!!

sorry i made a mistake. for my answer i should have written 7000(x - 10) instead of 7000(10 -x).

From your explanation i think the two answers are equivalent. The question i'm solving requires me to equate the derivative to 0 and solve for x. i asked about this specific differentiation because i thought that was where i was going wrong. but i think i need help with solving for x. my work is as follows:

3x/√(x²+9) = 7(10-x)/√(125-20x +x²)
√[(125-20x +x²) / (x²+9)] = 7(10-x) / 3x
(125-20x +x²) / (x²+9) = 49(10 -x)² / 9x²
9x²(125-20x +x²⁾ = 49(x²+9)(10 -x)²
1125x² - 180x³ + 9x⁴ = 49x⁴ - 980x³ + 5341x² - 8820x + 44100
0 = 40x⁴ - 800x³ + 4216x² - 8820x + 44100

How do I solve for x? Is there a simpler way?

 

[Steven G]

sorry i made a mistake. for my answer i should have written 7000(x - 10) instead of 7000(10 -x).

From your explanation i think the two answers are equivalent. The question i'm solving requires me to equate the derivative to 0 and solve for x. i asked about this specific differentiation because i thought that was where i was going wrong. but i think i need help with solving for x. my work is as follows:

3x/√(x²+9) = 7(10-x)/√(125-20x +x²)
√[(125-20x +x²) / (x²+9)] = 7(10-x) / 3x
(125-20x +x²) / (x²+9) = 49(10 -x)² / 9x²
9x²(125-20x +x²⁾ = 49(x²+9)(10 -x)²
1125x² - 180x³ + 9x⁴ = 49x⁴ - 980x³ + 5341x² - 8820x + 44100
0 = 40x⁴ - 800x³ + 4216x² - 8820x + 44100

How do I solve for x? Is there a simpler way?

Assuming you multiplied 49(x²+9)(10 -x)², your work is correct. I'd use the rational root test theorem which you learned in precalculus.