Question

find nth derivative of x^4/(x-1)(x-2)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given is a function $f(x)=\frac{x^4}{(x-1)(x-2)}$, Find its nth derivative.

Explanation:

• For a function of the form, $f(x)=log_e(ax+b)$ that is, the logarithm of the base 'e' form.
• Then the nth derivative of such a function 'f' is given by, $f^n(x)=\frac{(-1)^{n-1}(n-1)!a^n}{(ax+b)^n}$        -----(a)
• Here, we have $f(x)=y=\frac{x^4}{(x-1)(x-2)}$  
• Taking 'log' on both the sides of the above function we get,      
• $log(y)= log(\frac{x^4}{(x-1)(x-2)})\\log(y)=log(x^4)-log(x-1)-log(x-2)\\log(y)=4log(x)-log(x-1)-log(x-2)$  
• Using (a) we get the nth derivative for the above function as,
• $(\frac{(-1)^{n-1}(n-1)!}{y^n})y''^{..n}= 4(\frac{(-1)^{n-1}(n-1)!}{x^n})-(\frac{(-1)^{n-1}(n-1)!}{(x-1)^n})-(\frac{(-1)^{n-1}(n-1)!}{(x-2)^n}) \\y''^{..n}= y^n (\frac{4}{x^n} -\frac{1}{(x-1)^n} -\frac{1}{(x-2)^n} )\\f^n(x)= \frac{x^{4n}}{(x-1)^n(x-2)^n} (\frac{4}{x^n} -\frac{1}{(x-1)^n} -\frac{1}{(x-2)^n} )$      
• Hence the nth derivative of the given function is  $f^n(x)= \frac{x^{4n}}{(x-1)^n(x-2)^n} (\frac{4}{x^n} -\frac{1}{(x-1)^n} -\frac{1}{(x-2)^n} )$