Question

Let f(x) = ln(x^3 – 3x^2 – 16x – 12) , then the 1729^th derivative at x = 234 is?

(a) (1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\)

(b) (-1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\)

(c) (1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1728}}+\frac{1}{235^{1729}})\)

(d) (-1729!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\)

I got this question in exam.

My query is from The nth Derivative of Some Elementary Functions in chapter Differential Calculus of Engineering Mathematics

Answer 1

The correct option is (a) (1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\)

The best I can explain: The function can be written as f(x) = ln((x – 6)(x + 1)(x + 2))

Using property of logarithms we have

f(x) = ln(x + 1) + ln(x + 2) + ln(x – 6)

Using the n^th derivative of logarithmic functions

\(\frac{d^n(ln(x+a))}{dx^n}=\frac{(-1)^{n+1}\times(n-1)!}{(x+a)^n}\)

We have the 1729^th  derivative of f(x)  as

  \(f^{(1729)} (x) = \frac{1728!}{(x-6)^{1729}} + \frac{1728!}{(x+2)^{1729}} + \frac{1728!}{(x+1)^{1729}}\)

Now substituting x=234 we get our answer as

 = (1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\).