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Mathematics 21 Online
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

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

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):

\[(f^2)'=2f^{2-1}f'=2ff'\]

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 \[\frac{(1)(\frac{1}{5})+(2)(-5)}{\sqrt{(1)^2+(2)^2}}\]

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?

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

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

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?

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

myininaya (myininaya):

where are you trying to copy and paste from?

OpenStudy (nilankshi):

down

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