A P. answered • 03/18/20
Tutor
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Chemical Engineer with 10+ years of calculus teaching experience
y=(x^2/4)^x
y'=ln(x^2/4)*(x^2/4)^x*(x/2)=0
ln(x^2/4)*(x^3/8)=0
ln(x^2/4)=0
1=x^2/4
x^2=4, x= +/-2, so if it is only defined for x>=0, then x=2
Tommy G.
Implicitly derivate*
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03/18/20
A P.
tutor
You shouldn't need to do an implicit derivation (It doesn't really help you). I'm also challenging the +2 since I'm not sure exactly where it came from - the chain rule on x^2/4 is 2x/4=x/2, which should be multiplied. Does that make sense?
Let me know if you want to jump on a call.
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03/18/20
A P.
tutor
Ah, okay, slightly different method than I was anticipating. Should get the same answer. Let's see. With your method, implicit makes sense.
1/y dy/dx=ln(x^2/4) + 2
dy/dx=(ln(x^2/4)+2)y= 0 - This is where you are. I didn't bother to substitute the y in because you can divide it out immediately as follows. It's looking for horizontal tangents, aka where dy/dx or y'=0.
(ln(x^2/4)+2)y= 0
(ln(x^2/4)+2)= 0
(ln(x^2/4)= -2
e^-2=x^2/4
x^2=4e^-2
x=sqrt(4e^-2)
Hopefully this makes sense. Little tough to show on a computer.
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03/18/20
Tommy G.
It does make sense. Honestly, I also came up with the same x value but for some reason I keep doubting it. Thank you for your time & help❤.
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03/18/20
Tommy G.
Okay here is what I got when I implicitly dedicated the function. y ' = Y ln(x^2/4) + 2 0 = (x^2/4)^x (ln(x^2/4) + 2) So how do I solve for X03/18/20