Question

Sa se calculeze derivata functiei f apoi f'(x indice 0) .
a) f( x) = x/ x²+1 , x indice 0 = -1

b) f(x) = 2 sin x/ 3 cos x -5 , x indice 0 = 0

c) f(x) = ln x +x² / ln x-x² , x> 0

d) f( x) = x³-2x / x²+x+1 , x indice 0 = 1​

@ andyilye

Answer 1

Explicație pas cu pas:

a)

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

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

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

=>

$f'(-1) = \frac{1 - {( - 1)}^{2}}{( {( - 1)}^{2} + 1)^{2}} = \frac{1 - 1}{4} = \frac{0}{4} = 0$

b)

$f(x) = \frac{2 \sin(x) }{3 \cos(x) - 5}$

$f'(x) = \left(\frac{2 \sin(x) }{3 \cos(x) - 5} \right)' =$

$= 2 \cdot \frac{( \sin(x))' (3 \cos(x) - 5) - \sin(x) (3 \cos(x) - 5)'}{ {(3 \cos(x) - 5)}^{2} }$

$= 2 \cdot \frac{ \cos(x) (3 \cos(x) - 5) - \sin(x) ( - 3 \sin(x))}{ {(3 \cos(x) - 5)}^{2} }$

$= 2 \cdot \frac{ 3 \cos^{2} (x) - 5 \cos(x) + 3 \sin^{2} (x)}{ {(3 \cos(x) - 5)}^{2} }$

$= 2 \cdot \frac{ 3 \cos^{2} (x) - 5 \cos(x) + 3 \sin^{2} (x)}{ {(3 \cos(x) - 5)}^{2} }$

$= \frac{2(3 - 5 \cos(x)) }{{(3 \cos(x) - 5)}^{2} }$

=>

$f'(0) = \frac{2(3 - 5 \cos(0)) }{{(3 \cos(0) - 5)}^{2} } = \frac{2(3 - 5)}{ {(3 - 5)}^{2} } = \frac{ - 4}{4} = - 1$

c)

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

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

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

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

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

d)

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

=>

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

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

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

$= \frac{ {x}^{4} + 2x^{3} +5{x}^{2} -2}{( {x}^{2} + x + 1)^{2}}$

=>

$f'(1) = \frac{ {1}^{4} + 2 \times 1^{3} +5 \times {1}^{2} -2}{( {1}^{2} + 1 + 1)^{2}} = \frac{6}{9} = \frac{2}{3}$