Answer to Question #178216 in Calculus for zaz

Question #178216

Use logarithmic differentiation to find 𝑓 β€² (π‘₯) for 𝑓(π‘₯) = (π‘₯^2βˆ’1)(π‘₯ 2βˆ’2)(x^2βˆ’3) / (π‘₯^2+1)(π‘₯^2+2)(π‘₯^2+3) . Then, find 𝑓 β€² (2) and 𝑓 β€² (1).


1
Expert's answer
2021-04-15T07:33:59-0400

We have given that


"f(x) = \\dfrac{(\ud835\udc65^2\u22121)(\ud835\udc65^2\u22122)(x^2\u22123)}{(\ud835\udc65^2+1)(\ud835\udc65^2+2)(\ud835\udc65^2+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{(\ud835\udc65^2\u22121)(\ud835\udc65^2\u22122)(x^2\u22123)}{(\ud835\udc65^2+1)(\ud835\udc65^2+2)(\ud835\udc65^2+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}{(\ud835\udc65^2+1)(\ud835\udc65^2+2)(\ud835\udc65^2+3)}[ {2x(x^2-2)(x^2-3)}+{2x(x^2-1)(x^2-3)}+ {2x(x^2-1)(x^2-2)}- \\dfrac{2x{(\ud835\udc65^2\u22121)(\ud835\udc65^2\u22122)(x^2\u22123)}}{x^2+1}- \\dfrac{2x{(\ud835\udc65^2\u22121)(\ud835\udc65^2\u22122)(x^2\u22123)}}{x^2+2}- \\dfrac{2x{(\ud835\udc65^2\u22121)(\ud835\udc65^2\u22122)(x^2\u22123)}}{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\u22121)(1^2\u22122)(1^2\u22123)}}{1^2+1}- \\dfrac{2(1){(1^2\u22121)(1^2\u22122)(1^2\u22123)}}{1^2+2}- \\dfrac{2(1){(1^2\u22121)(1^2\u22122)(1^2\u22123)}}{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\u22121)(2^2\u22122)(2^2\u22123)}}{2^2+1}- \\dfrac{2(2){(2^2\u22121)(2^2\u22122)(2^2\u22123)}}{2^2+2}- \\dfrac{2(2){(2^2\u22121)(2^2\u22122)(2^2\u22123)}}{2^2+3}]"


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


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


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