Question

If the function $f\left( x \right)$ defined by $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1$, then $f'\left( 0 \right) = $
A. $100f'\left( 0 \right)$
B. $1$
C. $100$
D. None of these

Answer 1

Hint: In this problem, the polynomial function $f\left( x \right)$ is given. The degree of $f\left( x \right)$ is $100$ because the highest power of variable $x$ is $100$ in the first term. First, we will find differentiation of the given function $f\left( x \right)$ by using power rule. Power rule for differentiation is $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$. To find $f'\left( 0 \right)$, we will substitute $x = 0$ in differentiation of $f\left( x \right)$.


Complete step-by-step answer:
In this problem, it is given that $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1 \cdots \cdots \left( 1 \right)$. Note that here $f\left( x \right)$ is a polynomial function and the degree of $f\left( x \right)$ is $100$.
To find the value of $f'\left( 0 \right)$, first we will find differentiation of the given function $f\left( x \right)$ by using the power rule. Power rule for differentiation is $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$. Therefore, $f'\left( x \right) = \dfrac{{100\left( {{x^{99}}} \right)}}{{100}} + \dfrac{{99\left( {{x^{98}}} \right)}}{{99}} + ..... + \dfrac{{2x}}{2} + 1 + 0 \cdots \cdots \left( 2 \right)$
By cancellation of equal terms from numerator and denominator, we get
$f'\left( x \right) = {x^{99}} + {x^{98}} + ..... + x + 1 \cdots \cdots \left( 3 \right)$. Note that in equation $\left( 3 \right)$ every term of $f'\left( x \right)$ is containing variable $x$ except one constant term. Therefore, if we put $x = 0$ in equation $\left( 3 \right)$ then all those terms will be zero which contain variable $x$.
Now we are going to put $x = 0$ on both sides of equation $\left( 3 \right)$. Therefore, we get$f'\left( 0 \right) = {\left( 0 \right)^{99}} + {\left( 0 \right)^{98}} + ..... + \left( 0 \right) + 1$
$ \Rightarrow f'\left( 0 \right) = 1$
Therefore, if the function $f\left( x \right)$ defined by $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1$ then $f'\left( 0 \right) = 1$. Therefore, option B is correct.


Note: A polynomial function of $n$ degree in one variable $x$ is given by $f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ..... + {a_2}{x^2} + {a_1}x + {a_0}$ where ${a_n},{a_{n - 1}},.....,{a_1},{a_0}$ are real coefficients, $n$ is positive integer and ${a_n} \ne 0$. Every polynomial function is differentiable everywhere. Every differentiable function is continuous. So, we can say that every polynomial function is continuous. Power rule for derivatives is useful when we need to find the derivative of a variable raised to a power. The ${n^{th}}$ derivative of the function ${x^n}$ is $n!$.