Error in textbook answer?

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EJE12345

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f(x,y,z) = ln( (1-sqrt (x^2 + y^2 + z^3) / (1+sqrt (x^2 + y^2 + z^3) )

compute the partial derivative with respect to y at (1,2,2).

book gives 1/6 as the answer, but it doesn't look like (1,2,2) is in the domain of f.... f(1,2,2) is ln(-2) (only Real numbers allowed in this setting)

what am I missing?

(James Stewart, 8th Ed Calculus: Early Transcendentals, 14.3 prob 43)
 
EJE12345 said:
f(x,y,z) = ln( (1-sqrt (x^2 + y^2 + z^3) / (1+sqrt (x^2 + y^2 + z^3) )

compute the partial derivative with respect to y at (1,2,2).

book gives 1/6 as the answer, but it doesn't look like (1,2,2) is in the domain of f.... f(1,2,2) is ln(-2) (only Real numbers allowed in this setting)

what am I missing?

(James Stewart, 8th Ed Calculus: Early Transcendentals, 14.3 prob 43)
Click to expand...
I've typeset your formula using the forum's LaTeX (see https://www.freemathhelp.com/forum/threads/latex-math-formatting-is-available.109843/). Is this how it looks in the book ?:

[imath]f(x,y,z) = \ln\left(1-\frac{\sqrt {x^2 + y^2 + z^3}}{1+\sqrt {x^2 + y^2 + z^3} }\right)[/imath]
 
no, it is (1-...)/(1+...) (where the ... is the stuff under the square root)
 
[math]f(x,y,z) = \ln\left(\frac{1- \sqrt {x^2 + y^2 + z^3}}{1+\sqrt {x^2 + y^2 + z^3} }\right)[/math]
 
EJE12345 said:
f(x,y,z) = ln( (1-sqrt (x^2 + y^2 + z^3) / (1+sqrt (x^2 + y^2 + z^3) )
compute the partial derivative with respect to y at (1,2,2).
book gives 1/6 as the answer, but it doesn't look like (1,2,2) is in the domain of f.... f(1,2,2) is ln(-2) (only Real numbers allowed in this setting)
Click to expand...
Look at the partial derivative HERE.
If you have copied the question correctly note that at [imath](1,2,2)[/imath] we have:
[imath]1-\dfrac{\sqrt{x^2+y^2+z^3}}{1+\sqrt{x^2+y^2+z^3}}>0[/imath]
 
BigBeachBananas said:
[math]f(x,y,z) = \ln\left(\frac{1- \sqrt {x^2 + y^2 + z^3}}{1+\sqrt {x^2 + y^2 + z^3} }\right)[/math]
Click to expand...
yes, thats it, thanks. i guess the question i have if you're computing a partial derivative, you need the point to be in the domain, and it doesnt look like (1,2,2) is in the domain
 
Could the function be:

\(\displaystyle f(x,y,z) = \ln\left|\frac{1- \sqrt {x^2 + y^2 + z^3}}{1+\sqrt {x^2 + y^2 + z^3} } \right |\)

Where the vertical lines (||) will indicate absolute value.
 
maybe it should have absolute values, but the books doesn't have any.

also, i found 2 solutions on line that compute out the derivative and give that answer, but make no mention on the point not being in the domain. which makes me think im missing something (not assuming there is a typo)
 
[math]f_y=\dfrac{2y}{\left(y^2+z^3+x^2-1\right)\sqrt{y^2+z^3+x^2}}[/math]Then:[math]f_y(1,2,2)=\dfrac{2*2}{\left(2^2+2^3+1^2-1\right)\sqrt{2^2+2^3+1^2}}=\frac{1}{3\sqrt{13}}[/math]
 
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BigBeachBananas said:
[math]f_y=\dfrac{2y}{\left(y^2+z^3+x^2-1\right)\sqrt{y^2+z^3+x^2}}[/math]Then:[math]f_y(1,2,2)=\dfrac{2*2}{\left(2^2+2^3+1^2-1\right)\sqrt{2^2+2^3+1^2}}=\frac{1}{3\sqrt{13}}[/math]
Click to expand...
And if it actually uses [imath]z^2[/imath] instead of [imath]z^3[/imath] then you get your 1/6 for the answer.
 
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sorry, was supposed to be z^2, i didnt notice my typo. (in both numerator and denominator). apologies.

last question then - f(1,2,2) computes out to ln( (1-3)/(1+3) ) = ln(-0.5), correct? (sqrt (1^2 + 2^2 + 2^2) = sqrt(9) = 3)

if f(1,2,2) is ln(-0.5) then it is not a real number, so the partial derivative with respect to any of x,y,z wouldn't make sense (again, just real numbers, i dont know how this would work with complex numbers) ive looked over this so many times, i feel like im missing something small (like the z^2 which is now corrected)
 
EJE12345 said:
sorry, was supposed to be z^2, i didnt notice my typo. (in both numerator and denominator). apologies.

last question then - f(1,2,2) computes out to ln( (1-3)/(1+3) ) = ln(-0.5), correct? (sqrt (1^2 + 2^2 + 2^2) = sqrt(9) = 3)

if f(1,2,2) is ln(-0.5) then it is not a real number, so the partial derivative with respect to any of x,y,z wouldn't make sense (again, just real numbers, i dont know how this would work with complex numbers) ive looked over this so many times, i feel like im missing something small (like the z^2 which is now corrected)
Click to expand...
You are absolutely correct that [imath]\ln(-0.5)[/imath] is a complex number. But it's partial derivative can still be real, i.e., a complex number with a zero imaginary component.
 
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