Question

If $f\left( 1 \right)=1,f'\left( 1 \right)=3$ then the value of derivative of $f\left( f\left( f\left( x \right) \right) \right)+{{\left( f\left( x \right) \right)}^{2}}$ at x = 1 is:
a. 9
b. 33
c. 12
d. 20

Answer 1

This question involves the concept of differentiation. In this question, we have to calculate the derivative of a function. We will assume that a function as H(x), then we will find the derivative of that function with respect to x, using some rules of derivation like,
Rule 1: $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right)$
Rule 2: $\dfrac{d}{dx}\left( f\left( x \right).g\left( x \right) \right)=f'\left( x \right).g\left( x \right)+g'\left( x \right).f\left( x \right)$
Then, we will put x = 1 in the equation of H’(x) and using the values given in the question, we will solve the equation and get the value of H’(1).

Complete step by step answer:
It is given in the question that if $f\left( 1 \right)=1,f'\left( 1 \right)=3$, we have been asked to find the value of derivative of $f\left( f\left( f\left( x \right) \right) \right)+{{\left( f\left( x \right) \right)}^{2}}$ at x = 1.
So, let us consider the given function, $f\left( f\left( f\left( x \right) \right) \right)+{{\left( f\left( x \right) \right)}^{2}}$ as H(x). So, we can write,
$H\left( x \right)=f\left( f\left( f\left( x \right) \right) \right)+{{\left( f\left( x \right) \right)}^{2}}$
So, we have to find H’(x) at x = 1 and let H’(1).
Now, we will calculate H’(x).
We can find the derivative of H(x), by considering each term of that function individually, which are $f\left( f\left( f\left( x \right) \right) \right)$ and ${{\left( f\left( x \right) \right)}^{2}}$ and differentiating them.
So, first we will find the derivative of $f\left( f\left( f\left( x \right) \right) \right)$, so we get,

& \dfrac{d}{dx}\left[ f\left( f\left( f\left( x \right) \right) \right) \right]=f'\left( f\left( f\left( x \right) \right) \right)\times \dfrac{d}{dx}f'f\left( x \right) \\\ & =f'\left( f\left( f\left( x \right) \right) \right).f'\left( f\left( x \right) \right).f'\left( x \right) \\\ \end{aligned}$$ Now, we will find the derivative of ${{\left( f\left( x \right) \right)}^{2}}$, so we get, $\begin{aligned} & \dfrac{d}{dx}{{\left( f\left( x \right) \right)}^{2}}=f\left( x \right).f'\left( x \right)+f'\left( x \right).f\left( x \right) \\\ & =2f\left( x \right).f'\left( x \right) \\\ \end{aligned}$ So, we can write the derivative of H(x), that is H’(x) by adding the derivatives of both the terms of H(x), so we get, $\begin{aligned} & \dfrac{d}{dx}\left( H\left( x \right) \right)=\dfrac{d}{dx}f\left( f\left( f\left( x \right) \right) \right)+\dfrac{d}{dx}{{\left( f\left( x \right) \right)}^{2}} \\\ & H'\left( x \right)=f'\left( f\left( f\left( x \right) \right) \right).f'\left( f\left( x \right) \right).f'\left( x \right)+2f\left( x \right).f'\left( x \right) \\\ \end{aligned}$ On putting x = 1, we get, $H'\left( 1 \right)=f'\left( f\left( f\left( 1 \right) \right) \right).f'\left( f\left( 1 \right) \right).f'\left( 1 \right)+2f\left( 1 \right).f'\left( 1 \right)$ Now, we have been given that $f\left( 1 \right)=1,f'\left( 1 \right)=3$, so on substituting these values in the above equation, we get, $H'\left( 1 \right)=f'\left( f\left( 1 \right) \right).f'\left( 1 \right).3+\left( 2\times 1\times 3 \right)$ Again putting $f\left( 1 \right)=1,f'\left( 1 \right)=3$, we get, $H'\left( 1 \right)=f'\left( 1 \right)\times 3\times 3+\left( 6 \right)$ Again putting $f'\left( 1 \right)=3$, we get, $\begin{aligned} & H'\left( 1 \right)=3\times 3\times 3+\left( 6 \right) \\\ & H'\left( 1 \right)=27+6 \\\ & H'\left( 1 \right)=33 \\\ \end{aligned}$ Therefore, we get the value of the derivative of $f\left( f\left( f\left( x \right) \right) \right)+{{\left( f\left( x \right) \right)}^{2}}$ at x = 1 as 33. **So, the correct answer is “Option B”.** **Note:** In this question, for derivation of function ${{\left( f\left( x \right) \right)}^{2}}$, we can also use the rule, that is, $\dfrac{d}{dx}{{\left( f\left( x \right) \right)}^{n}}=n.{{\left( f\left( x \right) \right)}^{n-1}}.f'\left( x \right)$ And for derivation of $f\left( f\left( f\left( x \right) \right) \right)$, we can do the following. Consider $f\left( f\left( x \right) \right)=g\left( x \right)$. Hence, we can write, H(x) as, $H\left( x \right)=f\left( g\left( x \right) \right)$ So, its derivative will be, $H'\left( x \right)=f'\left( g\left( x \right) \right).g'\left( x \right)$ And the derivative of g(x) will be, $g'\left( x \right)=f'\left( f\left( x \right) \right).f'\left( x \right)$. So, we will get, $H'\left( x \right)=f'\left( f\left( f\left( x \right) \right) \right).f'\left( f\left( x \right) \right).f'\left( x \right)$