Question

Use logarithmic differentiation to find 𝑓 ′ (𝑥) for 𝑓(𝑥) = (𝑥^2−1)(𝑥 2−2)(x^2−3) / (𝑥^2+1)(𝑥^2+2)(𝑥^2+3) . Then, find 𝑓 ′ (2) and 𝑓 ′ (1).

Answer 1

We have given that

$f(x) = \dfrac{(𝑥^2−1)(𝑥^2−2)(x^2−3)}{(𝑥^2+1)(𝑥^2+2)(𝑥^2+3)}$

Taking log on both sides

$log{f(x)} = log[(x^2-1)(x^2-2)(x^2-3)]-log[(x^2+1)(x^2+2)(x^2+3)]$

$log{f(x)} = log(x^2-1)+log(x^2-2)+log(x^2-3)-log(x^2+1)-log(x^2+2)-log(x^2+3)$

Differentiating both sides w.r.t 'x'

$\dfrac{1}{f(x)}f'(x) = \dfrac{2x}{x^2-1}+\dfrac{2x}{x^2-2}+ \dfrac{2x}{x^2-3}- \dfrac{2x}{x^2+1}- \dfrac{2x}{x^2+2}- \dfrac{2x}{x^2+3}$

$f'(x) =f(x)[ \dfrac{2x}{x^2-1}+\dfrac{2x}{x^2-2}+ \dfrac{2x}{x^2-3}- \dfrac{2x}{x^2+1}- \dfrac{2x}{x^2+2}- \dfrac{2x}{x^2+3}]$

$f'(x) =\dfrac{(𝑥^2−1)(𝑥^2−2)(x^2−3)}{(𝑥^2+1)(𝑥^2+2)(𝑥^2+3)}[ \dfrac{2x}{x^2-1}+\dfrac{2x}{x^2-2}+ \dfrac{2x}{x^2-3}- \dfrac{2x}{x^2+1}- \dfrac{2x}{x^2+2}- \dfrac{2x}{x^2+3}]$

$f'(x) =\dfrac{1}{(𝑥^2+1)(𝑥^2+2)(𝑥^2+3)}[ {2x(x^2-2)(x^2-3)}+{2x(x^2-1)(x^2-3)}+ {2x(x^2-1)(x^2-2)}- \dfrac{2x{(𝑥^2−1)(𝑥^2−2)(x^2−3)}}{x^2+1}- \dfrac{2x{(𝑥^2−1)(𝑥^2−2)(x^2−3)}}{x^2+2}- \dfrac{2x{(𝑥^2−1)(𝑥^2−2)(x^2−3)}}{x^2+3}]$

$f'(1) =\dfrac{1}{(1^2+1)(1^2+2)(1^2+3)}[ {2(1)(1^2-2)(1^2-3)}+{2(1)(1^2-1)(x^2-3)}+ {2(1)(1^2-1)(1^2-2)}- \dfrac{2(1){(1^2−1)(1^2−2)(1^2−3)}}{1^2+1}- \dfrac{2(1){(1^2−1)(1^2−2)(1^2−3)}}{1^2+2}- \dfrac{2(1){(1^2−1)(1^2−2)(1^2−3)}}{1^2+3}]$

Hence, $f'(1) = \dfrac{1}{6}$

$f'(x) =\dfrac{1}{(2^2+1)(2^2+2)(2^2+3)}[ {2(2)(2^2-2)(2^2-3)}+{2(2)(2^2-1)(2^2-3)}+ {2(2)(2^2-1)(2^2-2)}- \dfrac{2(2){(2^2−1)(2^2−2)(2^2−3)}}{2^2+1}- \dfrac{2(2){(2^2−1)(2^2−2)(2^2−3)}}{2^2+2}- \dfrac{2(2){(2^2−1)(2^2−2)(2^2−3)}}{2^2+3}]$

Hence, $f'(2) = 40-\dfrac{552}{35}$

$f'(2) = \dfrac{848}{7350}$