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Mathematics 20 Online
OpenStudy (aravindg):

find dy/dx y=sqrt x raised to root x raised again to root x until infinity find dy/dx y=sqrt x raised to root x raised again to root x until infinity @Mathematics

OpenStudy (aravindg):

help

OpenStudy (agreene):

\[\huge\frac{d}{dx}\sqrt{x}^{\sqrt{x}^{\sqrt{x}^{...}}}\] is that what you are asking?

OpenStudy (aravindg):

ya

OpenStudy (aravindg):

pls help its urgent

OpenStudy (jamesj):

I wrote you an answer to this yesterday I think.

OpenStudy (aravindg):

i didnt get full answer :(

OpenStudy (aravindg):

i am stuck

OpenStudy (aravindg):

y=root x^y

OpenStudy (aravindg):

log y =y log root x

OpenStudy (aravindg):

now wat?

OpenStudy (jamesj):

Here it is again: Let \( y = f(x) = \sqrt{x}^ {...} = \sqrt{x}^{y} \) then \( \ln y = y. \ln (\sqrt{x}) = 1/2 . y . \ln x \) hence y'/y = 1/2 ( y/x + y' ln x)

OpenStudy (aravindg):

.........urgent

OpenStudy (jamesj):

Now solve algebraically for y'.

OpenStudy (aravindg):

how did 1/2 yog came?

OpenStudy (jamesj):

sqrt(x) = x^1/2 hence ln ( sqrt(x)) = 1/2 . ln x

OpenStudy (jamesj):

because ln(a^b) = b . ln a

OpenStudy (aravindg):

then what is answer?

OpenStudy (aravindg):

pls help its urgent i hav no time

OpenStudy (jamesj):

c'mon. You can solve this equation for y' : y'/y = 1/2 ( y/x + y' ln x)

OpenStudy (jamesj):

hence y' (1/y - ln x) = y/(2x) and therefore y' = ...

OpenStudy (aravindg):

k how did y'/y came?

OpenStudy (jamesj):

Implicit differentiation: (d/dx) ln y = dy/dx . 1/y

OpenStudy (aravindg):

thx help at this too f y= [2/(sqrt a^2-b^2)] x tan inverse [sqrt (a-b/a+b)]tan 1/2 x] then show that dy/dx=1/(a+bcosx)

OpenStudy (aravindg):

..........

OpenStudy (jamesj):

This one is just a messy calculation. If you know how to differentiate tan x and arctan x, then you should know how to do this. Also you need to know that 1 + tan^2 a = sec^2 a

OpenStudy (jamesj):

and the chain rule of course.

OpenStudy (aravindg):

plz hlp someone adviced me trigonometric substitution pls help its urgent

OpenStudy (jamesj):

No, trig substitution is not the answer. Let A = [2/(sqrt a^2-b^2)] and B = sqrt (a-b/a+b) Then you want the derivative of y = A arctan(B tan (x/2)) Write y = A arctan u, where u = B tan(x/2) Then \[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{A}{1+u^2} \ . \ B/2.\ \sec^2( x/2) \]

OpenStudy (jamesj):

Now substitute back in for A, B and u and carefully calculate.

OpenStudy (aravindg):

plz do it its urgent plssssssss

OpenStudy (aravindg):

helo

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