Question

can anyone answer the question on derivative? can anyone answer the question on derivative? @Mathematics

Answer 1

whats the question?

Answer 2

i will try ...

Answer 3

u will?

Answer 4

if 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)

Answer 5

can u do it its urgent

Answer 6

its out of my league, try using globalchautari.com /////////this must help you somehow...

Answer 7

?wat is that for?

Answer 8

visit, globalchautari.com, there you will find box for entering mathematical equations........try using it....

Answer 9

to saruz: what is difference b/w http://www.wolframalpha.com and globalchautari.com?

Answer 10

y= [2/(sqrt a^2-b^2)] x tan inverse [sqrt (a-b/a+b)]tan 1/2 x]

Answer 11

you need to put more parenthesis: is it (a-b)/(a+b) or a-b/a+b as it is?

Answer 12

So it looks like \[y=\frac{2}{\sqrt{a^2-b^2}} x tan^{-1}\left(\sqrt{\frac{a-b}{a+b}tan\frac{x}{2}}\right)\]?

Answer 13

no tan 1/2x

Answer 14

otside square root

Answer 15

pls help urgent

Answer 16

\[y=\frac{2}{\sqrt{a^2-b^2}} x tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\right)tan\frac{1}{2x}\]?

Answer 17

x in top

Answer 18

and i doubt if middle is x or into

Answer 19

helpppppp

Answer 20

\[y=\frac{2}{\sqrt{a^2-b^2}} x tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\right)tan(\frac{1}{2}x)\]?

Answer 21

k

Answer 22

no

Answer 23

tan 1/2 x inside bracket of tan inverse

Answer 24

That is why you need to type it by yourself with enough parenthesis

Answer 25

\[y=\frac{2}{\sqrt{a^2-b^2}} x tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}tan(\frac{1}{2}x)\right)\]?

Answer 26

ya

Answer 27

i thinkk middle is into *

Answer 28

not x

Answer 29

What do you mean by 'into'? Division?

Answer 30

multiply

Answer 31

*tan inversse

Answer 32

understand

Answer 33

Right, once more for absolute clarity then \[y=\frac{2}{\sqrt{a^2-b^2}} \times tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}tan(\frac{1}{2}x)\right)\]

Answer 34

y a finalized plz giv answr

Answer 35

@:MarinaDL, there is no any differences with mathematical section , however, there are lots of differences that you have noticed yourself.....

Answer 36

to saruz: thanks for the answer. It is easier to ask for short summary from somebody experienced than to figure out it on my own.

Answer 37

Alright, we've got an arctan, a tan, and a multiplication. \[\frac{d}{dx}tan^{-1}(x) = \frac{1}{1+x^2}\]\[\frac{d}{dx}tan(x) = sec^2 x\]and\[\frac{d}{dx} ax = a\]

The first fraction is a constant (let's call it k for now) since it doesn't contain an x. The fraction inside the arctan is also a constant, let's call it m.
\[k = \frac{2}{\sqrt{a^2 - b^2}}\]
\[m=\frac{a-b}{a+b}\]

Now we can chain the remaining terms:
\[y = k \times tan^{-1}(u) \Rightarrow \frac{dy}{du} = \frac{k}{1+u^2}\]\[u = m \times tan(v) \Rightarrow \frac{du}{dv} = m\times sec^2v\]and\[v=\frac{1}{2}x \Rightarrow \frac{dv}{dx} = \frac{1}{2}\]
\[\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dv}\times\frac{dv}{dx}\]

Is all of that cool AravindG?

Answer 38

\[\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dv}\times\frac{dv}{dx}\]

Answer 39

The question is to 'show that \(\frac{dy}{dx}=\frac{1}{a+bcosx} \)', so that is hopefully the answer.

Answer 40

You should probably work through the actual algebra at the end, and make sure you understand why I did all that chaining.

Answer 41

x=sin^3t/sqrt(cos 2t),y=cos^3t/sqrt(cos 2t) at t= pi/6 find dy/dx

Answer 42

y does not seem to be a function of x, so dy/dx would be 0.

Answer 43

I guess you could do some substitution, \[x=\frac{sin^3t}{\sqrt{cos(2t)}}\Rightarrow \sqrt{cos(2t)}=\frac{sin^3t}{x}\]\[y=\frac{cos^3t}{\sqrt{cos(2t)}} \Rightarrow y=\frac{cos^3t}{\frac{sin^3t}{x}}=\frac{x cos^3t}{sin^3t}=\frac{x}{tan^3t}\]I think that the tan is constant with respect to x, so \[\frac{dy}{dx} = \frac{1}{tan^3t}\]

Answer 44

wow now find values of a.,b,c such that lim x tends to 0 ((x(a+bcosx)-c*sinx)/x^5)=1

Answer 45

you have a funny expression inside the arctan

Answer 46

arctan ( k * tan u )

Answer 47

I dunno much about limits I'm afraid!

Answer 48

wt abt this find derivativre tan inverse (5 a x/a^2-6 x^2)

Answer 49

You're going to have to use the chain rule, and the quotient rule on that one.

Answer 50

plz show u see its very very urgent

Answer 51

ok

Answer 52

let k = sqrt ( a-b) / sqrt ( a+ b )

Answer 53

y = 2k/ (a-b) arctan [ k tan (1/2x)]

Answer 54

y' = 2k/(a-b) * 1/ { 1 + ( ktan(1/2*x))^2}* sec^2 ( 1/2*x)*1/2

Answer 55

\[y=tan^{-1}\left(\frac{5ax}{a^2-6x^2}\right)\]
Rewrite it as
\[y=tan^{-1}(w) \Rightarrow \frac{dy}{dx} = \frac{1}{1+w^2}\]\[w=\frac{5ax}{a^2-6x^2}\Rightarrow \frac{dy}{dx}=\frac{v \times \frac{du}{dx} - u\times \frac{dv}{dx}}{v^2}\]
Where \(u = 5ax, v = a^2 - 6x^2\)
I'm sure you can handle it from there bro!

Answer 56

Sorry, that first dy/dx should be dy/dw.

Answer 57

And the second one is dw/dx.

Answer 58

y' = 2k/(a-b) * 1/ { 1 + ( ktan(1/2*x))^2}* sec^2 ( 1/2*x)*1/2
y' = 2k/(a-b) * 1/ (1 + k^2*tan^2(1/2*x)* 1/cos^2( 1/2*x)*1/2

Answer 59

find dy/dx y=(x+1)^2sqrt(x-1)/(x+4)^3 e^x

Answer 60

are you doing a different problem now ?

Answer 61

((x+1)^2sqrt(x-1))/((x+4)^3 e^x)

Answer 62

u cntinue i need al answers

Answer 63

y = 2k/ (a-b) arctan [ k tan (1/2x)]
y' = 2k/(a-b) * 1/ { 1 + ( ktan(1/2*x))^2}* k*sec^2 ( 1/2*x)*1/2

Answer 64

y' = 2k/(a-b) * 1/ (1 + k^2*tan^2(1/2*x)* k/cos^2( 1/2*x)*1/2
y' = k/(a-b) * 1/ [k*cos^2(1/2x) + k^3*sin^2(1/2*x)]

Answer 65

y' = 1/(a-b) * 1/ [ cos^2(1/2x) + k^2 sin^2 (1/2*x)

Answer 66

y' = 1/(a-b) * 1/ ( cos^2(1/2x) + k^2 [1 - cos^2 (1/2*x)] )

Answer 67

still there?

Answer 68

ok i got it , it took a while, wow