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Let's make the question more clearer

Given : $(a+b)e^{\frac{y}{x}}=x$ ..........................(1)

To Prove: $x^{3}\frac{d^{2}y}{dx^{2}}=[x\frac{dy}{dx}-y]^{2}$

Proof

Here differentiating eqn (1)

We get $(a+b)e^{\frac{y}{x}}[\frac{x\frac{dy}{dx}-y}{x^{2}}] =1$

Substitute from eqn (1) , we get

$x\frac{dy}{dx}-y=x$ .....................(2)

Diff it , we get

$x\frac{d^{2}y}{dx^{2}}=1\\\\<br/>Multiplying\;by\;x^{2}\\\\<br/>x^{3}\frac{d^{2}y}{dx^{2}}=x^{2}\\\\<br/>Subsitute\; the \;values\; from\;eqn\;(2)\\\\<br/>x^{3}\frac{d^{2}y}{dx^{2}}=[x\frac{dy}{dx}-y]^{2}$

A little mistake by sovanANNA

lim _ (x - pi/2) 2cosx / pi -2x = 1

Use L hspital rule , -2sinx / -2 = 1 (when u put x= pi/2)

(A) 6

Let L = IZ₁-Z₂I² + IZ₂-Z₃I²+IZ₃-Z₁I²

Apply prop ZZ(bar) = IZI²

L = (Z₁-Z₂)( 1/Z₁ - 1/Z₂) + (Z₂-Z₃)(1/Z₂ - 1/Z₃) + (Z₃-Z₁)(1/Z₃ - 1/Z₁)

L = 6 - (some terms)

Thus L cannot xceed 6

here

(1+p)x² + 2(1+p)x+(1+p)=0 is the given equation

reduce it to ..

(1+p) [x²+2x+1] =0

(1+p)(x+1)² =0

Thus this eqn is already a square for all values of p except -1

Thus p belongs to R - {-1}

Try AK Sharma!!

here

S₁ = 1 + 6+11+16+21+26+31+36+41+46+51+.............n terms

S₂= 3 + 7+11+15+19+23+27+31+35+39+43+47+51..........n terms

Let S be the series of the common terms

S = 11+ 31 + 51+...........+ a₁₀

You can see the Common diff is 20

Therefore a₁₀ = a+9d = 11+9(20)=11+180 = 191

find the n th derivative of the following in terms of r and theta

y = tan^-1(2x / 1- x²)

pls help Urgently rq.

here

-1<=cos (2x+1) <= 1

cos^-1(-1)<=2x+1<= cos^-1(1)

- 0 <= 2x+1 <= 0

-1 < = 2x < = -1

[ -1 ] / 2 < = x < = -1 /2

could u give me some names of books on DIFFERENTIAL CALCULUS and INTEGRAL CALCULUS

is it

m v³(3-2root(2)) / 4g

here is the solution!!

I 'm not able to use the editor man

i'll try to show u normally

I = ? (x+root(1+x²)) / [(x+root(1+x²))]^1-n dx where ? is integral sign

Put x+root(1+x²) = t...............(2)

differentiate u get

[x+root(1+x²)] dx= dt root(1+x²) ...........(1)

Solve (2)

1+x²= t²+x²-2tx

u get x= t²-1 / 2t

Put it in (1)

[(x+root(1+x²))] dx = (1+t²) / 2t dt

Now

I =1/2 ? (1+t²) / t^2-n dt

now it's easy man!!

solve it

I = 1/2 ? t^n-2 dt + 1/2? tⁿ dt

ans will be I = 1/2 [ t^n-1 / n-1 + tⁿ⁺¹/ n+1]

substitute t

Oh i made a mistake nice one , thanks

Is the answer

15 or 3

Is it

2(u-v)

It is basically called as a SLAP = da/dt

but i think it is of no use (practically)!!