IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neil_mahaseth

Differential Coefficient

sriram.a

i could understand your first & 4rth Q) only.Here is the 4rth solution

${cos}^{-1}{(\frac{x^2-y^2}{x^2+y^2})}=loga,$

$\frac{-1}{\sqrt{1-x^2}}{(\frac{2x-2yy^1)(x^2+y^2)-(2x+2yy^1)(x^2-y^2)}{(x^2+y^2)^2})}=0$ y^1=dy/dx

simplifie the above one,

The answer is y/x

ashish_banga

is the answer to second F(1) = 0

mujtaba4iit

See the solution for 4.

$\frac{x^2- y^2}{x^2+y^2}\,=cos{(log a)}=k\,\,\,\,(say)<br/>\\ By \,componendo-dividendo\\\,\frac{x^2}{-y^2}= \frac{k+1}{k-1}\\\,\frac{x^2}{y^2}= \frac{1+k}{1-k}\\\,\frac{x}{y}= \sqrt\frac{1+k}{1-k}\\\,Now\,differentiating\, wrt\,x\\\,\frac{d}{dx}{(\frac{x}{y})}=0\\\, y-x{\frac{dy}{dx}}=0\\\,y=x{\frac{dy}{dx}}\\\,so\, {\frac{dy}{dx}}=\frac{y}{x}$

Plz nudge me if u dont understand any step.

This sum can be done in different ways also, try by some other method.

mujtaba4iit

Plz nudge me if my answer to the first question is right, then i'll post the solution.

I got this answer:

$\frac{dy}{dx}=\frac{nxy-my^2}{nx(x+y)}\\$

Plz anyone nudge me if its correct.

studyid

2)

$y = (x+\sqrt{1+{x}^{2}})^{n}$

$Differentiating\;w.r.t\;x,\;we\;have$

${y}_{1} = n*(x+\sqrt{1+{x}^{2}})^{n-1}\frac{d}{dx}(x+\sqrt{1+{x}^{2}})$

$So\; {y}_{1} = n\frac{(x+\sqrt{1+{x}^{2}})^{n}}{(x+\sqrt{1+{x}^{2}})}(1+\frac{2x}{2\sqrt{1+{x}^{2}}})$

$So\; {y}_{1} = n\frac{(x+\sqrt{1+{x}^{2}})^{n}}{(x+\sqrt{1+{x}^{2}})}(\frac{(x+\sqrt{1+{x}^{2}})}{\sqrt{1+{x}^{2}}})$

$So\;(\sqrt{1+{x}^{2}}){y}_{1} = ny$

$Squaring\; Both \;Sides :(1+{x}^{2}){{y}_{1}}^{2} = {n}^{2}{y}^{2}$

$Again\;Differentiating\;w.r.t\;x(1+{x}^{2})2{y}_{1}{y}_{2} + 2x{{y}_{1}}^{2}={n}^{2}2y{y}_{1}$

$Taking\;out\;2{y}_{1}\;common$

$We \;Get\;(1+{x}^{2}){y}_{2} + x{y}_{1} = {n}^{2}y$

$Hence\;Final\;Answer\;is\;{n}^{2}y$

ziauddin75

ans for first one is y/x

ziauddin75

last waale ka ans aa raha hai 3aP(x) a is any arbitrary const ans sahii hai toh bahao aage method type karoon

sriram.a

for 5th)y^2=p(x)

where y^I=dy/dx,Y^II=d^2y/dx^2

$2yy^{II}+2(y^I)^2=P^I(x)$

mutiply both sides by y^2

$2y^3y^{II}+2(yy^I)^2=P^I(x)P(x)$ ,sorry p^I(x)P^II(x)

$2y^3y^{II}=P^I(x)P^{II}(x)-\frac{(P^I(x))^2}{2}$

$2\frac{d(y^3y^{II})}{dx}=P^I(x)P^{II}(x)+P^{III}(x)P^I(x)-P^I(x)P^{II}(x)$

$so,\frac{d(y^3\frac{d^2y}{dx^2}}{dx}=\frac{P^{III}(x)P(x)}{2}$

ziauddin75

my answer is also the same the fact is i had taken it really 3rd degree as
ax3+bx2+cx+d

mujtaba4iit

------------------------------------------------------------solution not correct hence deleted----------------------------------------------------------

ziauddin75

when yu differentiaited ist time it must be 1+dy/dx yu had written 1 only that all yu will get soln now

ashish_banga

second one

$IF(x) - F (y) I\le {(x-y)}^{2}$

$\frac{mod ( F(x) - F(y)}{x-y} \le (x-y)$

this gives

$mod (F'(x)) \le 0$

but mod is always greater than or equal to 0

so F'(x) = 0

F(x) is constant

so F (1) = 0

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