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