Question

Va rog dau coroana . Sa se caculeze derivatele . De la 1 la 7 si si de la 10 la 12​

Answer 1

Explicație pas cu pas:

1.

$f'(x) = (({x}^{2} - 3x)^{5})' \\ = 5({x}^{2} - 3x)^{4}({x}^{2} - 3x)' \\ = 5({x}^{2} - 3x)^{4}(2x - 3)$

2.

$f'(x) = (( \frac{x}{x + 1} )^{3})' \\ = 3( \frac{x}{x + 1} )^{2}(\frac{x}{x + 1})' \\ = 3( \frac{x}{x + 1} )^{2}(\frac{1}{(x + 1)^{2} }) \\ = \frac{3{x}^{2}}{(x + 1)^{4}}$

3.

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

4.

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

5.

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

6.

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

7.

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

10.

$f'(x) = (sin(3x+1))' \\ = cos(3x+1)(3x+1)' \\ = 3cos(3x+1)$

11.

$f'(x) = ( \frac{1}{2} cos({x}^{2} ))' \\ = \frac{1}{2} ( - \sin({x}^{2} ))({x}^{2})' \\ = \frac{1}{2} ( - \sin({x}^{2} )) \times 2x \\ = - x\sin({x}^{2})$

12.

$f'(x) = (tan({x}^{2} + 1))' \\ = \sec^{2} ({x}^{2} + 1) ( {x}^{2} + 1)' \\ = 2x\sec^{2} ({x}^{2} + 1)$