Question

If g(x) is the inverse function of $f\left( x \right)$ and $f'(x)=\dfrac{1}{1+{{x}^{4}}}$ , then $g'(x)$ is equal to:
(a) $1+{{\left( g(x) \right)}^{4}}$
(b) $1-{{\left( g(x) \right)}^{4}}$
(c) $1+{{\left( f(x) \right)}^{4}}$
(d) $\dfrac{1}{1+{{\left( g(x) \right)}^{4}}}$

Answer 1

Hint: We know that if $f(x)\text{ and g(x)}$ are inverse of each other then we know that $f\left( g(x) \right)=x$ . So, differentiate both sides of this equation using chain rule of differentiation. Finally use the definition if f’(x), i.e., $f'(x)=\dfrac{1}{1+{{x}^{4}}}$ and rearrange the equation to get the final answer.

Complete step by step solution:
As it is given that $f(x)\text{ and g(x)}$ are inverse of each other, we can say that $f\left( g(x) \right)=x$ . Now we will differentiate both sides of this equation.
$\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=\dfrac{dx}{dx}$
Now for the differentiation of left-hand side, we will use the chain rule of differentiation, according to which $\dfrac{d\left( f\left( g(x) \right) \right)}{dx}=f'\left( g(x) \right)\times g'(x)$ . So, our equation comes out to be:
$f'\left( g(x) \right)\times g'(x)=1......(i)$
Also, it is given in the question that $f'(x)=\dfrac{1}{1+{{x}^{4}}}$ . So, if we substitute x in the definition by g(x), we get
$f'(g(x))=\dfrac{1}{1+{{\left( g(x) \right)}^{4}}}$
If we substitute this value in equation (i), we get
$\dfrac{1}{1+{{\left( g(x) \right)}^{4}}}\times g'(x)=1$
$\Rightarrow g'(x)=1+{{\left( g(x) \right)}^{4}}$
Therefore, the answer to the above question is option (a).

Note: Remember that the inverse of a function represents the reflection of the original function about the y=x line when represented on the graph. Also, be very careful while differentiating using chain rule, as according to chain rule the outermost function is differentiated first and the innermost at the end. If you find your answer not matching with the options given in the questions when you solve by taking $f\left( g(x) \right)=x$ , try to solve using $g\left( f(x) \right)=x$ , as this is also a valid relation.