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
help
\[\huge\frac{d}{dx}\sqrt{x}^{\sqrt{x}^{\sqrt{x}^{...}}}\] is that what you are asking?
ya
pls help its urgent
I wrote you an answer to this yesterday I think.
i didnt get full answer :(
i am stuck
y=root x^y
log y =y log root x
now wat?
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)
.........urgent
Now solve algebraically for y'.
how did 1/2 yog came?
sqrt(x) = x^1/2 hence ln ( sqrt(x)) = 1/2 . ln x
because ln(a^b) = b . ln a
then what is answer?
pls help its urgent i hav no time
c'mon. You can solve this equation for y' : y'/y = 1/2 ( y/x + y' ln x)
hence y' (1/y - ln x) = y/(2x) and therefore y' = ...
k how did y'/y came?
Implicit differentiation: (d/dx) ln y = dy/dx . 1/y
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)
..........
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
and the chain rule of course.
plz hlp someone adviced me trigonometric substitution pls help its urgent
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 \[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{A}{1+u^2} \ . \ B/2.\ \sec^2( x/2) \]
Now substitute back in for A, B and u and carefully calculate.
plz do it its urgent plssssssss
helo
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