Question

let g(x) be the inverse of f(x) and
df/dx =1/1+logx
then value of
dg/dx - log(g(x)) is equal to=?

a) 0
b) -1
c) 1
d) g(x) +2​

Answer 1

Answer is given in the pic.

Answer 2

Given : function g(x) is the inverse of f(x) and

df/dx = 1/(1 + logx)

To find : dg/dx - log(g(x)) is equal to ...

solution : as it has given that, g(x) is inverse of f(x)

so, g(x) = f¯¹(x) ⇒g(f(x)) = x

now differentiating with respect to x we get,

g'(f(x)) f'(x) = 1

⇒g'(f(x)) df/dx = 1

given, df/dx = 1/(1 + logx)

so, g'(f(x)) × 1/(1 + logx) = 1

⇒g'(f(x)) = (1 + logx)

now let f(x) = y ⇒f¯¹(y) = x = g(y)

so, g'(y) = 1 + log{g(y)}

⇒g'(y) - log{g(y)} = 1

⇒g'(x) - log{g(x)} = 1 [ putting x = y ]

⇒dg/dx - log{g(x)} = 1

therefore dg/dx - log{g(x)} is equal to 1.