Mathematics
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OpenStudy (anonymous):
find the derivative of sqrt((f(x)^2+g(x)^2). when you know that f(x)=1 , g(x)=2, f(x)prime=1/5, and g(x)prime=-5
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myininaya (myininaya):
\[h(x)=\sqrt{[f(x)]^2+[g(x)]^2}\]
\[h'(x)=\frac{1}{2}(2f(x) f'(x)+2g(x)g'(x))^\frac{-1}{2}\]
myininaya (myininaya):
just plug in
myininaya (myininaya):
and i'm sure those aren't the values for all x
myininaya (myininaya):
this latex is not working for me :(
OpenStudy (anonymous):
ya we were given a table of f and g values and their derivatives
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OpenStudy (anonymous):
i love those problems. use the formula, plug in the numbers
OpenStudy (anonymous):
so i keep getting ((1/2)(Fx^2+gx^2)^(-1/2))*(2fx+gx)*(2fxprime +2gxprime) and idk waht im doing wrong
OpenStudy (anonymous):
its the chain rule right?
OpenStudy (anonymous):
when i put it in i get a nonreal answer and it will not take that
myininaya (myininaya):
\[(f^2)'=2f^{2-1}f'=2ff'\]
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OpenStudy (anonymous):
so im typing the answer in as 1/(2(sqrt((2*1*(1/5)+(2*2*-5)))
myininaya (myininaya):
\[(f^2+g^2)'=2ff'+2g g'\]
myininaya (myininaya):
\[[(f^2+g^2)^\frac{1}{2}]'=\frac{1}{2}[(f^2+g^2)^{\frac{1}{2}-1}](2ff'+2g g')\]
myininaya (myininaya):
\[=\frac{2ff'+2g g'}{2\sqrt{f^2+g^2}}=\frac{ff'+g g'}{\sqrt{f^2+g^2}}\]
myininaya (myininaya):
f=1,g=2,f'=1/5, g'=-5
\[\frac{(1)(\frac{1}{5})+(2)(-5)}{\sqrt{(1)^2+(2)^2}}\]
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myininaya (myininaya):
\[=\frac{\frac{1}{5}-10}{\sqrt{1+4}}\]
myininaya (myininaya):
\[=\frac{\frac{1-50}{5}}{\sqrt{5}}\]
OpenStudy (nilankshi):
good
OpenStudy (anonymous):
haha ok that worked ty
myininaya (myininaya):
=\[\frac{\frac{-49}{5}}{\sqrt{5}}\]
you got it?
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OpenStudy (anonymous):
but i dont really know why there is the part on the top and bottom
OpenStudy (anonymous):
ohhhh i got it lol
myininaya (myininaya):
what do you mean?
OpenStudy (anonymous):
chain rule right?
myininaya (myininaya):
yep
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OpenStudy (anonymous):
ty lol
myininaya (myininaya):
we used a bunch of chain rule
OpenStudy (anonymous):
ya i can see lol
myininaya (myininaya):
good! :)
OpenStudy (anonymous):
good rule that one
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OpenStudy (anonymous):
not so easy to prove though right?
myininaya (myininaya):
what nilankshi?
OpenStudy (anonymous):
bad copy and paste
OpenStudy (nilankshi):
therer is some prob with this page
myininaya (myininaya):
oh is that what that is?
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OpenStudy (anonymous):
maybe mathml run amok
OpenStudy (nilankshi):
i can many symbol there
myininaya (myininaya):
not again
myininaya (myininaya):
lol
OpenStudy (nilankshi):
like this what is this
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myininaya (myininaya):
where are you trying to copy and paste from?
OpenStudy (nilankshi):
down