Question

Let f (x) > 0 for all x and f’ (x) exists for all x. if f is the inverse function of h and $\left( {h'\left( x \right) = \dfrac{1}{{1 + \log x}}} \right)$, then f’ (x) will be ?
$\left( a \right)1 + \log f\left( x \right)$
$\left( b \right)1 + f\left( x \right)$
$\left( c \right)1 - \log f\left( x \right)$
$\left( d \right)\log f\left( x \right)$

Answer 1

In this particular question use the concept that inverse function is a function that reverses another function so if f is the inverse function of h, then h (f (x)) = x, then differentiate both sides w.r.t. x to calculate the value of f’ (x) in terms of h’ (f (x)), so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data: f (x) > 0 for all x and f’ (x) exists for all x
And f is the inverse function of h,
Therefore h (f (x)) = x
Now differentiate the above equation w.r.t x we have,
$\Rightarrow \dfrac{d}{{dx}}\left[ {h\left( {f\left( x \right)} \right)} \right] = \dfrac{d}{{dx}}x$
Now as we know that $\dfrac{d}{{dx}}u\left( {g\left( x \right)} \right) = u'g\left( x \right)\dfrac{d}{{dx}}g\left( x \right),\dfrac{d}{{dx}}x = 1$, so use this property in the above equation we have,
$\Rightarrow h'\left( {f\left( x \right)} \right)\dfrac{d}{{dx}}f\left( x \right) = 1$
$\Rightarrow h'\left( {f\left( x \right)} \right)f'\left( x \right) = 1$, $\left[ {\because \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right)} \right]$
$\Rightarrow f'\left( x \right) = \dfrac{1}{{h'\left( {f\left( x \right)} \right)}}$.................. (1)
Now it is given that $\left( {h'\left( x \right) = \dfrac{1}{{1 + \log x}}} \right)$
So in the above equation substitute f (x) in place of x we have,
$\Rightarrow h'\left( {f\left( x \right)} \right) = \dfrac{1}{{1 + \log f\left( x \right)}}$
Now substitute this value in equation (1) we have,
$\Rightarrow f'\left( x \right) = \dfrac{1}{{\dfrac{1}{{1 + \log f\left( x \right)}}}} = 1 + \log f\left( x \right)$
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation property which is given as $\dfrac{d}{{dx}}u\left( {g\left( x \right)} \right) = u'g\left( x \right)\dfrac{d}{{dx}}g\left( x \right),\dfrac{d}{{dx}}x = 1$, so differentiate the equation (1) by using this property and then substitute the values as above we will get the required value of f’(x).