Question

Let f(x) = ln(x^2 + 5x + 6) then the value of f^(30)(1) is given by?

(a) (29!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)

(b) (-29!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)

(c) (30!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)

(d) (-30!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)

I have been asked this question in semester exam.

Question is from The nth Derivative of Some Elementary Functions topic in portion Differential Calculus of Engineering Mathematics

Answer 1

Correct choice is (b) (-29!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)

Explanation: Given function

f(x) = ln(x^2 + 5x + 6)

Factorising the inner polynomial we get

f(x) = ln((x + 3) (x + 2))

Now using the rule of logarithms ln (m * n) = ln(m) + ln(n)

we get

f(x) = ln(x + 3) + ln(x + 2)

Now using the n^th derivative of logarithmic function \(\frac{d^n(ln(x+a))}{dx^n} = \frac{(-1)^{n+1}\times(n-1)!}{(x+a)^n}\)

We have

\(\frac{d^n(f(x))}{dx^n}=\frac{d^n(ln(x+3))}{dx^n}+\frac{d^n(ln(x+2))}{dx^n}\)

Which simplifies to \(\frac{d^n(f(x))}{dx^n}=\frac{(-1)^{n+1}\times(n-1)!}{(x+3)^n}+\frac{(-1)^{n+1}\times(n-1)!}{(x+2)^n}\)

Substituting x=1 and n=30

Gives us the answer as (-29!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)