Question

If $f(x)$ and $g(x)$ are two functions with $g(x) = x - \dfrac{1}{x}$ and $fog(x) = {x^3} - \dfrac{1}{{{x^3}}}$ , then find $f'(x)$ .
 A. $3{x^2} + 3$
 B. \[{x^2} - \dfrac{1}{x}\]
 C. $1 + \dfrac{1}{{{x^2}}}$
 D. $3{x^2} + \dfrac{3}{{{x^4}}}$

Answer 1

Hint: We have been given $g(x)$ and $fog(x)$ . To solve this question we first find out the value of $f(x)$ then differentiate it to find $f'(x)$ . Now to find $f(x)$ put $g(x)$ in $fog(x)$ and use arithmetic identities accordingly. Arithmetic identities used are: ${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$ and ${a^2} + {b^2} = {(a - b)^2} + 2ab$ . Finally after all the calculations to find $f(x)$, substitute $x - \dfrac{1}{x} = x$ .

Complete step by step solution: 
Given: $g(x) = x - \dfrac{1}{x}$ and $fog(x) = {x^3} - \dfrac{1}{{{x^3}}}$ ...(1)
We know that $fog(x) = f(g(x))$ (composition function)
This implies that $fog(x) = f(x - \dfrac{1}{x})$ ...(2)
This is because $g(x) = x - \dfrac{1}{x}$ .
Substituting equation (2) in equation (1), we get
 $f(x - \dfrac{1}{x}) = {x^3} - \dfrac{1}{{{x^3}}}$
Now we use the identity, ${a^3} - {b^3} = (a - b)({a^2} + {b^2} + ab)$ where $a = x$ and $b = \dfrac{1}{x}$ .
Therefore, $f(x - \dfrac{1}{x}) = \left( {x - \dfrac{1}{x}} \right)\left( {{x^2} + {{\left( {\dfrac{1}{x}} \right)}^2} + \left( {x \times \dfrac{1}{x}} \right)} \right)$
 $f(x - \dfrac{1}{x}) = \left( {x - \dfrac{1}{x}} \right)\left( {{x^2} + {{\left( {\dfrac{1}{x}} \right)}^2} + 1} \right)$
Now again we use an arithmetic identity, ${a^2} + {b^2} = {(a - b)^2} + 2ab$ where $a = x$ and $b = \dfrac{1}{x}$ .
We get, \[f(x - \dfrac{1}{x}) = \left( {x - \dfrac{1}{x}} \right)\left( {{{\left( {x - \dfrac{1}{x}} \right)}^2} + 2\left( {x \times \dfrac{1}{x}} \right) + 1} \right)\]
This implies that, \[f(x - \dfrac{1}{x}) = \left( {x - \dfrac{1}{x}} \right)\left( {{{\left( {x - \dfrac{1}{x}} \right)}^2} + 2 + 1} \right)\]
\[f(x - \dfrac{1}{x}) = \left( {x - \dfrac{1}{x}} \right)\left( {{{\left( {x - \dfrac{1}{x}} \right)}^2} + 3} \right)\] ...(3)
Now if we substitute $x - \dfrac{1}{x} = x$ , we can get the value of $f(x)$ and after differentiating the value of $f(x)$ we can finally get the value of $f'(x)$ which is asked in the question.
So, now we put $x - \dfrac{1}{x} = x$ in equation number (3), we get
 \[f(x) = x({x^2} + 3)\]
Open the brackets, we get
 \[f(x) = {x^3} + 3x\] ...(4)
Now that we have obtained the value of $f(x)$ , we will now find the value of $f'(x)$ .
Differentiating both the sides of equation number (4), we get
 $f'(x) = 3{x^2} + 3$
This is because the differentiation of ${x^n}$ is ${x^{n - 1}}$ .
Hence, the correct option is A. $3{x^2} + 3$ .

Note: To solve this question one must have the knowledge of composition functions and its identities. Also, knowing arithmetic identities is important for such questions. These questions can also be asked in trigonometric terms so knowing concepts and identities of trigonometry is important here. The last step to find $f(x)$ should be understood carefully.