Question

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Answer 1

f(x)=ln((x²+4)/(x²-1));f'(2)=?

f'(x)=(1:(x²+4)/(x²-1))*((x²+4)/(x²-1))'=

(x²-1)/(x²+4)*(2x(x²-1)-2x(x²+4))/(x²-1)²=

(1/(x²+4))*2x(x²-1-x²-4)/(x²-1)=

-10x/((x²+4)(x²-1))

f'(2)=-20/(8*3)==-5/6

1)(lnf(x))'=(1/(f(x)))*f'(x)

2)(u/v)'=(u'•v-v'•u)/v²

Answer 2

$f(x)=ln\frac{x^{2}+4 }{x^{2}-1 }\\\\f'(x)=(ln\frac{x^{2}+4 }{x^{2}-1 })'=\frac{1}{\frac{x^{2}+4 }{x^{2}-1 }}*(\frac{x^{2}+4 }{x^{2}-1 })'=\frac{x^{2}-1 }{x^{2}+4}*\frac{(x^{2}+4)'*(x^{2}-1)-(x^{2}+4)*(x^{2}-1)'}{(x^{2}-1)^{2}}=\frac{x^{2}-1 }{x^{2}+4 }*\frac{2x(x^{2}-1)-2x(x^{2}+4)}{(x^{2}-1)^{2}}=\frac{x^{2}-1 }{x^{2}+4 }*\frac{2x(x^{2}-1-x^{2}-4)}{(x^{2}-1)^{2}}=\frac{2x*(-5)}{(x^{2}+4)(x^{2}-1)}=-\frac{10x}{(x^{2}+4)(x^{2}-1)}\\\\f'(2)=-\frac{10*2}{(2^{2}+4)(2^{2}-1)}=-\frac{20}{8*3}=-\frac{5}{6}$

Использована формула :

$(lnx)'=\frac{1}{x}$