Differentiation - 5 - Inverse Trigonometric Functions

d/dx of sin^-1x = 1/√(1-x²)

d/dx of cos^-1x = -1/√(1-x²)

d/dx of tan^-1x = 1/(1+x²)

d/dx of cot^-1x = -1/(1+x²)

d/dx of sec^-1x = 1/|x|√(x²-1)

d/dx of cosec^-1x = -1/|x|√(x²-1)

Differentiation - Model Problems - 6

Problems having some complexity and requiring use multiple concepts from this chapter as well as other chapters of algebra, trigonometry and geometry

1. Prob:

If y = tan^-1[(√(1+x²) -1)/x]
find y'(1)

To solve the problem, transform the given expression by putting x = tan θ or θ = tan^-1x

[(√(1+x²) -1)/x] = [(√(1+tan²θ) - 1)/tan θ)
=> (√(sec²θ)-1)/tan θ = (sec θ - 1)/tan θ

writing sec θ = 1/cos θ and tan θ = sin θ/cos θ

(sec θ - 1)/tan θ = (1- cos θ)/sin θ
=> tan θ/2

Therefore y tan^-1(tan θ/2) = θ/2
=> y = (tan^-1x)/2

As d/dx of tan^-1x = 1(1+x²)

dy/dx = (1/2)* 1(1+x²)

Therefore y'(1) = (1/2)* 1(1+1) = (1/2)*(1/2) = 1/4



2. Prob: if f(x) = log_x(ln x), then find value of f'(x) at e. (JEE 1985)

Solution:
f(x) = log_x(ln x),
=> x^f(x) = ln(x)

Taking logaritm on both sides
f(x)ln x = ln (ln x)
Write y = f(x)
y (ln x) = ln (ln x)

Differentiating both sides with respect to x

y*(1/x)+(ln x)*(dy/dx) = (1/ln x)* (1/x)

substitute x = e

y*(1/e) + (ln e)*(dy/dx) = (1/ln e)*(1/e)
=> dy/dx = (1/e) - (y/e)

value of y at e = log_x(ln e) = log_x(1) = 0

Therefore dy/dx = 1/e

Differentiation - Model Problems - 7

1. There is exists a function f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x , and

a. f’’(x)>0 for all x
b. -1c. -2≤f’’(x) ≤-1 or all x
d. f’’(x)<-2 for all x

(JEE, 1982)

Answer: (a)

Reason:

x² -x+1 = (x – ½)² + ¾ can be a solution to the f(x) satisfying f(0) = 1, f’(0) = -1, f(x)>0 for all x as f(0) = 1 and f’(0) = -1

f’’(x) = 2 and hence f’’(x)>0 for all x.

We may assume f(x) = e^-x and get a similar conclusion. But e^-x may tend to zero as x tends to infinity. Hence f(x) = x² -x+1 is more appropriate function.

2. If y = f[(2x-1)/( x²+1)] and f’(x) = sin x², then dy/dx = --------------.
(JEE 1982)

Answer: dy/dx = sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²

Solution: f’(x) = sin x²
=> f’(t) = sin t²
Given problem is visualized as
y = f(t)
t = [(2x-1)/( x²+1)]
dy/dx = dy/dt*dt/dx
=> sin t² [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [( x²+1)(2) – (2x-1)(2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2x²+2 –(4x²-2x)]/ ( x²+1) ²
=> sin [(2x-1)/( x²+1)] ² * [2 +2x–2x²]/ ( x²+1) ²