Question

Sa se calculeze derivata functiei f apoi f' (x indice 0) .

a) f(x) = 1/ x-1 , x indice 0 = 2

b) f(x) = 1/ x²-3x+2 , x indice 0 = 0

c) f(x) = ln x +x / ln x -x , x >0 , x indice 0 = 1

d) f(x) = radical x / x²+1 , x >sau egal 0 ​

Va rog dau coroana urgent

Answer 1

Explicație pas cu pas:

a)

$f(x) = \frac{1}{x - 1}$

=>

$f'(x) = \left(\frac{1}{x - 1}\right)' = -\frac{(x - 1)'}{(x - 1)^{2}} = - \frac{1}{(x - 1)^{2}}$

=>

$f'(2) = - \frac{1}{ {(2 - 1)}^{2} } = - \frac{1}{1} = -1$

.

b)

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

=>

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

=>

$f'(0) = \frac{3 - 2\cdot 0}{({0}^{2} - 3\cdot 0 + 2)^{2}} = \frac{3}{(2)^{2}} = \frac{3}{4}$

.

c)

$f(x) = \frac{ ln(x) + x }{ln(x) - x}$

=>

$f'(x) = \left(\frac{ ln(x) + x}{ln(x) - x}\right)' =$

$= \frac{\left(ln(x) + x \right)' \left(ln(x) - x \right) - \left(ln(x) + x \right) \left(ln(x) - x \right)'}{\left(ln(x) - x \right)^{2}}$

$= \frac{\left( \frac{1}{x} + 1 \right) \left(ln(x) - x \right) - \left(ln(x) + x \right) \left( \frac{1}{x} - 1 \right)}{\left(ln(x) - x \right)^{2}}$

$= \frac{2(ln(x) - 1)}{\left(ln(x) - x \right)^{2}}$

=>

$f'(1) = \frac{2(ln(1) - 1)}{(ln(1) - 1)^{2}} = \frac{2(0 - 1)}{(0 - 1)^{2}} = -2$

.

d)

$f(x) = \frac{ \sqrt{x} }{x^{2} + 1}$

=>

$f'(x) = \left(\frac{ \sqrt{x} }{x^{2} + 1}\right)'$

$= \frac{( \sqrt{x} )'( {x}^{2} + 1) - \sqrt{x}( {x}^{2} + 1)' }{ {({x}^{2} + 1)}^{2} }$

$= \frac{\left( \frac{1}{2 \sqrt{x} } \right)( {x}^{2} + 1) - 2x\sqrt{x}}{ {({x}^{2} + 1)}^{2} }$

$= \frac{1 - 3 {x}^{2} }{ 2 \sqrt{x} {({x}^{2} + 1)}^{2} }$