If y = a_0 + a_1x + a_2x^2 + dots + a_nx^n,th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the polynomial function $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$. To find the $n^{th}$ derivative $y_n$,we differentiate $y$ with respect

Vedclass · Accepted Answer

$n!a_n$

Vedclass · Answer

(C) Given the polynomial function $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$. To find the $n^{th}$ derivative $y_n$,we differentiate $y$ with respect to $x$ repeatedly $n$ times. The first derivative is $y_1 = \frac{dy}{dx} = a_1 + 2a_2x + 3a_3x^2 + \dots + na_nx^{n-1}$. The second derivative is $y_2 = \frac{d^2y}{dx^2} = 2 	imes 1 a_2 + 3 	imes 2 a_3x + \dots + n(n-1)a_nx^{n-2}$. Continuing this process,the $k^{th}$ derivative of $x^n$ is given by $\frac{d^k}{dx^k}(x^n) = \frac{n!}{(n-k)!}x^{n-k}$. For $k=n$,the $n^{th}$ derivative of $x^n$ is $\frac{n!}{(n-n)!}x^{n-n} = n!$. All terms with powers of $x$ less than $n$ will become zero after $n$ differentiations. Therefore,$y_n = \frac{d^n}{dx^n}(a_nx^n) = a_n 	imes n!$.

If $y = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$,then find the $n^{th}$ derivative $y_n$.

Similar Questions

If $f(x) = \frac{x^2}{x+a}$,then $f^{\prime \prime}(a)$ is equal to

If $y = \sin^{-1} x$,then $(1 - x^2) y_2 - x y_1 = $

Find the second order derivative of the function $f(x) = x^{2} + 3x + 2$.

$A$ non-zero polynomial with real coefficients has the property that $f''(x) f'(x) = f(x)$. Then the value of $f'''(x)$ is

If $y=ax^{n+1}+b x^{-n}$,then $x^2 \frac{d^2 y}{d x^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module