Question

1
1. Find the nth derivative of
1/1-5x+6x2​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given,

f(x) = $\frac{1}{1-5x+6x^2}$

To find,

The nth derivative of the function f(x)

Solution,

We can simply solve the problem by the following process.

By the partial fraction method of integrals, f(x) can be written in the following way.

f(x) = $\frac{1}{(2x-1)(3x-1)}$

     = $\frac{A}{2x-1} + \frac{B}{3x-1}$

To find the value of A;

2x-1 = 0

x = $\frac{1}{2}$

A = $\frac{1}{3x-1}$

   = $\frac{1}{3*\frac{1}{2} -1}$

   = 2

In the same way, to find the value of B;

3x - 1 = 0

x = $\frac{1}{3}$

B = $\frac{1}{2x-1}$  (Putting the value of x)

  = -3

Thus,

f(x) = $\frac{2}{2x-1} - \frac{3}{3x-1}$

Now,

To find the nth derivative of f(x), we must know that ;

nth derivative of (ax+ b)⁻¹ is $\frac{(-1)^n n! a^n}{(ax+b)^(1+n)}$.

Thus,

nth derivative of f(x) = $\frac{(-1)^n n! 2^n}{(2x-1)^(1+n)} - \frac{(-1)^n n! 3^n}{(3x-1)^(1+n)}$

Thus, the above is the answer to the question.