Test: Second Order Derivatives - JEE MCQ

y = tan^-1 x
=> tan y = tan(tan^-1 x)
x = tan y
sec² y dy/dx = 1
=> 1/(cos²y) dy/dx = 1
dy/dx = cos²y
Differentiate it with respect to x
d²y/dx² = 2cos y(-sin y)(dy/dx)..................(1)
Put the value of dy/dx in eq(1)
=> d²y/dx² = 2cos y(-sin y)(cos² y)
=> d²y/dx² = 2cos³ y siny

x^a y^b = (x−y)^a+b
taking log both the sides.
log(x^a y^b)=log(x−y)^(a+b)
alogx+blogy=(a+b)log(x−y)
differentiating both sides w.r.t 'x'.
(a/x+b/y)dy/dx = (a+b)/(x−y)[1−dy/dx]
dy/dx[b/y+(a+b)/(x−y)] = (a+b)/(x−y) − a/x
dy/dx[(bx−by+ay+by)/y(x−y)] = (ax+bx−ax+ay)/x(x−y)
dy/dx[(bx+ay)/y]=(bx+ay)/x
dy/dx = y/x

 f(x) = x + cot x
f’(x) = 1 + (-cosec² x)
f”(x) = 0 - 2cosec x(-cosec x cot x)
= 2 cosec² x cot x
f”(π/4) = 2 cosec² (π/4) cot(π/4)
= 2 [(2)^½]² (1)
= 4

The operator form of given equation is (x²D² – 5xD + 9)y = 0 ...(*)
Let x = e^t ⇒ t = log x D′ ≡ d / dt
D ≡ d / dt
We have x²D² = D′(D′ – 1) 
xD = D′ 
[D′(D′ – 1) – 5D′ + 9]y = 0 
The A.E. is m2 – 6m + 9 = 0 
(m – 3)² = 0, m = 3, 3 
The C.F. is y = (c¹ + c²t)e^3t 
The solution of (*) is 
y = (c₁ + c₂ log x) x³.

e^y =1/ (x+5)
Taking log both side we get
log e^y = log {1/(5+x)}
then, y = log {1/(5+x)}
Differentiating both side ,
dy/dx =( x+5) . {-1/(5+x)²}
dy/dx = -1/(5+x) ……..( 1)
Again Differentiating, d²y/dx²= 1/(5+x)²
d²y/dx²= {-1/(5+x)}²
From equation (1)
d²y/dx² = {dy/dx}²

y = at⁴,     x = 2at²
dy/dt = 4at³        dx/dt = 4at    => dt/dx = 1/4at
Divide dy/dt by dx/dt, we get
dy/dx = t²
d² y/dx² = 2t dt/dx……………….(1)
Put the value of dt/dx in eq(1)
d² y /dx² = 2t(1/4at)
= 1/2a