2.0
f ( x ) = x 4 − x + 1 {\displaystyle f\left(x\right)\ =\ x^{4}-x+1}
f ( 1 ) = 1 4 − 1 + 1 = 1 ⇒ ( 1 , 1 ) {\displaystyle f\left(1\right)\ =\ 1^{4}-1+1=1\Rightarrow \ \left(1,1\right)}
f ( x + h ) = ( x + h ) 4 − x + h + 1 {\displaystyle f\left(x+h\right)=\left(x+h\right)^{4}-x+h+1}
f ( x + h ) − f ( x ) = ( x + h ) 4 − x + h + 1 − ( x 4 − x + 1 ) = ( x + h ) 4 − ( x 4 + h ) = {\displaystyle f\left(x+h\right)-f\left(x\right)=\left(x+h\right)^{4}-x+h+1-\left(x^{4}-x+1\right)=\left(x+h\right)^{4}-\left(x^{4}+h\right)=}
x 4 + 4 x 3 h + 6 x 2 h 2 + 4 x h 3 + h 4 − x 4 − h = 4 x 3 h + 6 x 2 h 2 + 4 x h 3 + h 4 − h {\displaystyle x^{4}+4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}-x^{4}-h=4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}-h}
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 4 x 3 h + 6 x 2 h 2 + 4 x h 3 + h 4 − h h = {\displaystyle f'(x)=\lim _{h\to 0}{\frac {f\left(x+h\right)-f\left(x\right)}{h}}=\lim _{h\to 0}{\frac {4x^{3}h+6x^{2}h^{2}+4xh^{3}+h^{4}-h}{h}}=}
lim h → 0 h ( 4 x 3 + 6 x 2 h + 4 x h 2 + h 3 − 1 ) h = lim h → 0 4 x 3 + 6 x 2 h + 4 x h 2 + h 3 − 1 = 4 x 3 − 1 {\displaystyle \lim _{h\to 0}{\frac {h\left(4x^{3}+6x^{2}h+4xh^{2}+h^{3}-1\right)}{h}}=\lim _{h\to 0}4x^{3}+6x^{2}h+4xh^{2}+h^{3}-1=4x^{3}-1}
f ′ ( 1 ) = 4 ⋅ ( 1 ) 3 − 1 = 3 {\displaystyle f'\left(1\right)=4\cdot \left(1\right)^{3}-1=3}
y − 1 = 3 ( x − 1 ) {\displaystyle y-1=3\left(x-1\right)}
y = 3 x − 2 {\displaystyle y=3x-2}
3.1
f ( x ) = ( ( ( 2 x + 3 ) 4 + 5 ) 6 + 7 ) 8 {\displaystyle f\left(x\right)=\left(\left(\left(2x+3\right)^{4}+5\right)^{6}+7\right)^{8}}
f ′ ( x ) = 8 ⋅ ( ( ( 2 x + 3 ) 4 + 5 ) 6 + 7 ) 7 ⋅ 6 ( ( 2 x + 3 ) 4 + 5 ) 5 ⋅ 4 ( 2 x + 3 ) 3 ⋅ 2 = {\displaystyle f'\left(x\right)=8\cdot \left(\left(\left(2x+3\right)^{4}+5\right)^{6}+7\right)^{7}\cdot 6\left(\left(2x+3\right)^{4}+5\right)^{5}\cdot 4\left(2x+3\right)^{3}\cdot 2=}
384 ⋅ ( ( ( 2 x + 3 ) 4 + 5 ) 6 + 7 ) 7 ⋅ ( ( 2 x + 3 ) 4 + 5 ) 5 ⋅ ( 2 x + 3 ) 3 {\displaystyle 384\cdot \left(\left(\left(2x+3\right)^{4}+5\right)^{6}+7\right)^{7}\cdot \left(\left(2x+3\right)^{4}+5\right)^{5}\cdot \left(2x+3\right)^{3}}
3.2
f ( x ) = ( ( x + 1 ) 2 + 1 ) 2 ( x 2 − 3 x − 3 ) 20 {\displaystyle f\left(x\right)={\frac {\left(\left(x+1\right)^{2}+1\right)^{2}}{\left(x^{2}-3x-3\right)^{20}}}}
u = ( ( x + 1 ) 2 + 1 ) 2 = ( x 2 + 2 x + 1 + 1 ) 2 = ( x 2 + 2 x + 2 ) 2 {\displaystyle u=\left(\left(x+1\right)^{2}+1\right)^{2}=\left(x^{2}+2x+1+1\right)^{2}=\left(x^{2}+2x+2\right)^{2}}
v = ( x 2 − 3 x − 3 ) 20 {\displaystyle v=(x^{2}-3x-3)^{20}}
u ′ = 2 ( x 2 + 2 x + 2 ) ( 2 x + 2 ) = ( 4 x 2 + 8 x + 8 ) ( x + 1 ) = 4 x 3 + 8 x 2 + 8 x + 4 x 2 + 8 x + 8 = 4 x 3 + 12 x 2 + 16 x + 8 {\displaystyle u'=2\left(x^{2}+2x+2\right)\left(2x+2\right)=\left(4x^{2}+8x+8\right)\left(x+1\right)=4x^{3}+8x^{2}+8x+4x^{2}+8x+8=4x^{3}+12x^{2}+16x+8}
v ′ = 20 ( x 2 − 3 x − 3 ) 19 ⋅ ( 2 x − 3 ) {\displaystyle v'=20\left(x^{2}-3x-3\right)^{19}\cdot \left(2x-3\right)}
f ′ ( x ) = ( 4 x 3 + 12 x 2 + 16 x + 8 ) ⋅ ( x 2 − 3 x − 3 ) 20 − 20 ( x 2 − 3 x − 3 ) 19 ⋅ ( 2 x − 3 ) ⋅ ( x 2 + 2 x + 2 ) 2 ( ( x 2 − 3 x − 3 ) 20 ) 2 = {\displaystyle f'(x)={\frac {\left(4x^{3}+12x^{2}+16x+8\right)\cdot \left(x^{2}-3x-3\right)^{20}-20\left(x^{2}-3x-3\right)^{19}\cdot \left(2x-3\right)\cdot \left(x^{2}+2x+2\right)^{2}}{\left(\left(x^{2}-3x-3\right)^{20}\right)^{2}}}=}
( x 2 − 3 x − 3 ) 19 ( ( 4 x 3 + 12 x 2 + 16 x + 8 ) ( x 2 − 3 x − 3 ) − 20 ( 2 x − 3 ) ( x 2 + 2 x + 2 ) 2 ) ( x 2 − 3 x − 3 ) 40 = {\displaystyle {\frac {(x^{2}-3x-3)^{19}\left(\left(4x^{3}+12x^{2}+16x+8\right)\left(x^{2}-3x-3\right)-20\left(2x-3\right)\left(x^{2}+2x+2\right)^{2}\right)}{(x^{2}-3x-3)^{40}}}=}
( 4 x 3 + 12 x 2 + 16 x + 8 ) ( x 2 − 3 x − 3 ) − 20 ( 2 x − 3 ) ( x 2 + 2 x + 2 ) 2 ( x 2 − 3 x − 3 ) 21 = {\displaystyle {\frac {(4x^{3}+12x^{2}+16x+8)(x^{2}-3x-3)-20(2x-3)(x^{2}+2x+2)^{2}}{(x^{2}-3x-3)^{21}}}=}
4 x 5 − 12 x 4 − 12 x 3 + 12 x 4 − 36 x 3 − 36 x 2 + 16 x 3 − 48 x 2 − 48 x + 8 x 2 − 24 x − 24 − 40 x 3 − 20 x 2 + 40 x + 120 ( x 2 − 3 x − 3 ) 21 = {\displaystyle {\frac {4x^{5}-12x^{4}-12x^{3}+12x^{4}-36x^{3}-36x^{2}+16x^{3}-48x^{2}-48x+8x^{2}-24x-24-40x^{3}-20x^{2}+40x+120}{(x^{2}-3x-3)^{21}}}=}
4 x 5 − 72 x 3 − 96 x 2 − 32 x + 96 ( x 2 − 3 x − 3 ) 21 {\displaystyle {\frac {4x^{5}-72x^{3}-96x^{2}-32x+96}{\left(x^{2}-3x-3\right)^{21}}}}
3.3
f ( x ) = l n ( l n ( 3 x − 12 x 2 + 1 ) ) {\displaystyle f(x)=ln\left(ln\left({\frac {3x-12}{x^{2}+1}}\right)\right)}
f ′ ( x ) = 1 l n ( 3 x − 12 x 2 + 1 ) ⋅ x 2 + 1 3 x − 12 ⋅ 3 ( x 2 + 1 ) − 2 x ⋅ 3 ( x − 4 ) ( x 2 + 1 ) 2 {\displaystyle f'(x)={\frac {1}{ln\left({\frac {3x-12}{x^{2}+1}}\right)}}\cdot {\frac {x^{2}+1}{3x-12}}\cdot {\frac {3\left(x^{2}+1\right)-2x\cdot 3\left(x-4\right)}{\left(x^{2}+1\right)^{2}}}}
f ′ ( x ) = ( x 2 + 1 ) − 2 x ⋅ ( x − 4 ) l n ( 3 x − 12 x 2 + 1 ) ⋅ ( x 2 + 1 ) ⋅ ( x − 4 ) {\displaystyle f'\left(x\right)={\frac {\left(x^{2}+1\right)-2x\cdot \left(x-4\right)}{ln\left({\frac {3x-12}{x^{2}+1}}\right)\cdot \left(x^{2}+1\right)\cdot (x-4)}}}
f ′ ( x ) = − x 2 + 8 x + 1 ln ( 3 x − 12 x 2 + 1 ) ( x − 4 ) ( x 2 + 1 ) {\displaystyle f'(x)={\frac {-x^{2}+8x+1}{\ln \left({\frac {3x-12}{x^{2}+1}}\right)\left(x-4\right)\left(x^{2}+1\right)}}}
3.4
f ( x ) = 7 x 2 − 3 6 {\displaystyle f(x)={\sqrt[{6}]{7x^{2}-3}}}
f ′ ( x ) = 1 6 ⋅ 1 ( 7 x 2 − 3 ) 5 6 ⋅ 2 ⋅ 7 x = 7 x 3 ⋅ ( 7 x 2 − 3 ) 5 6 {\displaystyle f'(x)={\frac {1}{6}}\cdot {\frac {1}{\sqrt[{6}]{\left(7x^{2}-3\right)^{5}}}}\cdot 2\cdot 7x={\frac {7x}{3\cdot {\sqrt[{6}]{\left(7x^{2}-3\right)^{5}}}}}}
3.5
f ( x ) = tan ( 5 x + x 2 − 2 x − 1 ) {\displaystyle f(x)=\tan \left(5x+{\sqrt {x^{2}-2x-1}}\right)}
f ′ ( x ) = 5 + 1 2 x 2 − 2 x − 1 ⋅ ( 2 x − 2 ) cos 2 ( 5 x + x 2 − 2 x − 1 ) {\displaystyle f'(x)={\frac {5+{\frac {1}{2{\sqrt {x^{2}-2x-1}}}}\cdot \left(2x-2\right)}{\cos ^{2}\left(5x+{\sqrt {x^{2}-2x-1}}\right)}}}
f ′ ( x ) = 10 x 2 − 2 x − 1 + ( 2 x − 2 ) 2 x 2 − 2 x − 1 cos 2 ( 5 x + x 2 − 2 x − 1 ) 1 {\displaystyle f'(x)={\frac {\frac {10{\sqrt {x^{2}-2x-1}}+\left(2x-2\right)}{2{\sqrt {x^{2}-2x-1}}}}{\frac {\cos ^{2}\left(5x+{\sqrt {x^{2}-2x-1}}\right)}{1}}}}
f ′ ( x ) = 10 x 2 − 2 x − 1 + ( 2 x − 2 ) 2 x 2 − 2 x − 1 ⋅ cos 2 ( 5 x + x 2 − 2 x − 1 ) {\displaystyle f'(x)={\frac {10{\sqrt {x^{2}-2x-1}}+\left(2x-2\right)}{2{\sqrt {x^{2}-2x-1}}\cdot \cos ^{2}\left(5x+{\sqrt {x^{2}-2x-1}}\right)}}}
f ′ ( x ) = 5 x 2 − 2 x − 1 + x − 1 x 2 − 2 x − 1 ⋅ cos 2 ( 5 x + x 2 − 2 x − 1 ) {\displaystyle f'(x)={\frac {5{\sqrt {x^{2}-2x-1}}+x-1}{{\sqrt {x^{2}-2x-1}}\cdot \cos ^{2}\left(5x+{\sqrt {x^{2}-2x-1}}\right)}}}
3.6
f ( x ) = ( x − 5 ) 2 x + 4 = e l n ( x − 5 ) 2 x + 4 = e ( 2 x + 4 ) ⋅ l n ( x − 5 ) {\displaystyle f(x)=(x-5)^{2x+4}=e^{ln(x-5)^{2x+4}}=e^{(2x+4)\cdot ln(x-5)}}
f ′ ( x ) = e ( 2 x + 4 ) ⋅ l n ( x − 5 ) ⋅ ( 2 l n ( x − 5 ) + 2 x + 4 x − 5 ) = ( x − 5 ) 2 x + 4 ⋅ ( 2 l n ( x − 5 ) + 2 x + 4 x − 5 ) {\displaystyle f'(x)=e^{(2x+4)\cdot ln(x-5)}\cdot (2ln(x-5)+{\frac {2x+4}{x-5}})=\left(x-5\right)^{2x+4}\cdot \left(2ln\left(x-5\right)+{\frac {2x+4}{x-5}}\right)}
3.7
f ( x ) = sin x cos x = e ln ( sin x cos x ) = e cos x ⋅ ln ( sin x ) {\displaystyle f\left(x\right)=\sin x^{\cos x}=e^{\ln \left(\sin x^{\cos x}\right)}=e^{\cos x\cdot \ln \left(\sin x\right)}}
f ′ ( x ) = e cos x ⋅ ln ( sin x ) ⋅ ( − sin x ⋅ ln ( sin x ) + 1 sin x ⋅ cos x ⋅ cos x ) = sin x cos x ⋅ ( − sin x ⋅ ln ( sin x ) + cos 2 x sin x ) {\displaystyle f'\left(x\right)\ =e^{\cos x\cdot \ln \left(\sin x\right)}\cdot \left(-\sin x\cdot \ln \left(\sin x\right)+{\frac {1}{\sin x}}\cdot \cos x\cdot \cos x\right)=\sin x^{\cos x}\cdot \left(-\sin x\cdot \ln \left(\sin x\right)+{\frac {\cos ^{2}x}{\sin x}}\right)}
3.8
f ( x ) = sin ( x cos x ) = sin ( e ln ( x cos x ) ) = sin ( e cos x ⋅ ln x ) {\displaystyle f\left(x\right)=\sin \left(x^{\cos x}\right)=\sin \left(e^{\ln \left(x^{\cos x}\right)}\right)=\sin \left(e^{\cos x\cdot \ln x}\right)}
f ′ ( x ) = cos ( e cos x ⋅ ln x ) ⋅ e cos x ⋅ ln x ⋅ ( − sin x ⋅ ln x + cos x x ) = cos ( x cos x ) ⋅ x cos x ⋅ ( − sin x ⋅ ln x + cos x x ) {\displaystyle f'\left(x\right)=\cos \left(e^{\cos x\cdot \ln x}\right)\cdot e^{\cos x\cdot \ln x}\cdot \left(-\sin x\cdot \ln x+{\frac {\cos x}{x}}\right)=\cos \left(x^{\cos x}\right)\cdot x^{\cos x}\cdot \left(-\sin x\cdot \ln x+{\frac {\cos x}{x}}\right)}
4.1
f ( x ) = e e x {\displaystyle f\left(x\right)=e^{e^{x}}}
f ′ ( x ) = e e x ⋅ e x {\displaystyle f'\left(x\right)=e^{e^{x}}\cdot e^{x}}
f ′ ( x ) = e e x ⋅ e x ⋅ e x + e x ⋅ e e x = e x ⋅ e e x ( e x + 1 ) = e e x + x ( e x + 1 ) {\displaystyle f'\left(x\right)=e^{e^{x}}\cdot e^{x}\cdot e^{x}+e^{x}\cdot e^{e^{x}}=e^{x}\cdot e^{e^{x}}\left(e^{x}+1\right)=e^{e^{x}+x}\left(e^{x}+1\right)}
4.2
y = 3 e ( 2 x + 5 ) + 16 x + 4 + sin ( 2 x ) {\displaystyle y=3e^{\left(2x+5\right)}+16x+4+\sin \left(2x\right)}
y ′ = 6 e ( 2 x + 5 ) + 16 + 2 cos ( 2 x ) {\displaystyle y'=6e^{\left(2x+5\right)}+16+2\cos \left(2x\right)}
y ″ = 12 e ( 2 x + 5 ) − 4 sin ( 2 x ) {\displaystyle y''=12e^{\left(2x+5\right)}-4\sin \left(2x\right)}
y ‴ = 24 e ( 2 x + 5 ) − 8 cos ( 2 x ) {\displaystyle y'''=24e^{\left(2x+5\right)}-8\cos \left(2x\right)}
y ⁗ = 48 e ( 2 x + 5 ) + 16 sin ( 2 x ) {\displaystyle y''''=48e^{\left(2x+5\right)}+16\sin \left(2x\right)}
y ′′′′′ = 96 e ( 2 x + 5 ) + 32 cos ( 2 x ) {\displaystyle y'''''=96e^{\left(2x+5\right)}+32\cos \left(2x\right)}
y ′′′′′′ = 192 e ( 2 x + 5 ) − 64 sin ( 2 x ) {\displaystyle y''''''=192e^{\left(2x+5\right)}-64\sin \left(2x\right)}
y ′′′′′′′ = 384 e ( 2 x + 5 ) − 128 cos ( 2 x ) {\displaystyle y'''''''=384e^{\left(2x+5\right)}-128\cos \left(2x\right)}
![{\displaystyle f(x)={\sqrt[{6}]{7x^{2}-3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/651f83b82693bf0cd2e41ef95adf567e288b83b5)
![{\displaystyle f'(x)={\frac {1}{6}}\cdot {\frac {1}{\sqrt[{6}]{\left(7x^{2}-3\right)^{5}}}}\cdot 2\cdot 7x={\frac {7x}{3\cdot {\sqrt[{6}]{\left(7x^{2}-3\right)^{5}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09d94ee5582221db5d9ef37b426205050888da3b)
