f ′ ( x ) = D ( x ) N ′ ( x ) − N ( x ) D ′ ( x ) D 2 ( x ) {\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}} N ′ ( x ) = d g d h d h d x = 3 ( 2 x + 4 ) 2 ( 2 ) = 6 ( 2 x + 4 ) 2 {\displaystyle N'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=3(2x+4)^{2}(2)=6(2x+4)^{2}} D ′ ( x ) = 12 x 2 {\displaystyle D'(x)=12x^{2}}
f ′ ( x ) = A ′ ( x ) B ( x ) + A ( x ) B ′ ( x ) {\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)} A ′ ( x ) = 2 {\displaystyle A'(x)=2} B ′ ( x ) = d g d h d h d x = 1 2 2 x + 2 ( 2 ) = 1 2 x + 2 {\displaystyle B'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}}
f ′ ( x ) = D ( x ) N ′ ( x ) − N ( x ) D ′ ( x ) D 2 ( x ) {\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}} N ′ ( x ) = 2 {\displaystyle N'(x)=2} D ′ ( x ) = d g d h d h d x = 1 2 2 x + 2 ( 2 ) = 1 2 x + 2 {\displaystyle D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}}
f ′ ( x ) = A ′ ( x ) B ( x ) + A ( x ) B ′ ( x ) {\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)} A ′ ( x ) = d g d h d h d x = 1 2 2 x 2 + 1 ( 4 x ) = 2 x 2 x 2 + 1 {\displaystyle A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x^{2}+1}}}}(4x)={\frac {2x}{\sqrt {2x^{2}+1}}}} B ′ ( x ) = d r d s d s d x = 2 ( 3 x 4 + 2 x ) ( 12 x 3 + 2 ) {\displaystyle B'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=2(3x^{4}+2x)(12x^{3}+2)}
f ′ ( x ) = D ( x ) N ′ ( x ) − N ( x ) D ′ ( x ) D 2 ( x ) {\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}} N ′ ( x ) = 2 {\displaystyle N'(x)=2} D ′ ( x ) = d g d h d h d x = 2 ( x 4 + 4 x + 2 ) ( 4 x 3 + 4 ) {\displaystyle D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x^{4}+4x+2)(4x^{3}+4)}
f ′ ( x ) = A ′ ( x ) B ( x ) + A ( x ) B ′ ( x ) {\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)} A ′ ( x ) = d g d h d h d x = 1 2 x 3 + 1 ( 3 x ) = 3 x 2 x 3 + 1 {\displaystyle A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {x^{3}+1}}}}(3x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}} B ′ ( x ) = 2 x {\displaystyle B'(x)=2x}
f ′ ( x ) = d g d h d h d x = 2 ( A ( x ) + B ( x ) + 2 ) ( A ′ ( x ) + B ′ ( x ) ) {\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(A(x)+B(x)+2)(A'(x)+B'(x))} A ′ ( x ) = d r d s d s d x = 4 ( 2 x + 3 ) 3 ( 2 ) = 8 ( 2 x + 3 ) 3 {\displaystyle A'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=4(2x+3)^{3}(2)=8(2x+3)^{3}} B ′ ( x ) = 8 {\displaystyle B'(x)=8}
那么
使用链式法则,我们得到
各个因子为
因此
f ′ ( x ) = A ′ ( x ) B ( x ) C ( x ) + A ( x ) B ′ ( x ) C ( x ) + A ( x ) B ( x ) C ′ ( x ) {\displaystyle f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)} A ′ ( x ) = B ′ ( x ) = C ′ ( x ) = 1 {\displaystyle A'(x)=B'(x)=C'(x)=1}
使用隐函数求导法求 y'
3 y 2 y ′ − x y ′ = y − 3 x 2 {\displaystyle 3y^{2}y'-xy'=y-3x^{2}}
8 ( 2 x + y ) 3 + 4 y ′ ( 2 x + y ) 3 + 6 x + 6 y y ′ = 1 y − x y ′ y 2 {\displaystyle 8(2x+y)^{3}+4y'(2x+y)^{3}+6x+6yy'={\frac {1}{y}}-{\frac {xy'}{y^{2}}}} y ′ ( 4 ( 2 x + y ) 3 + 6 y + x y 2 ) = 1 y − 8 ( 2 x + y ) 3 − 6 x {\displaystyle y'(4(2x+y)^{3}+6y+{\frac {x}{y^{2}}})={\frac {1}{y}}-8(2x+y)^{3}-6x} y ′ ( 4 y 2 ( 2 x + y ) 3 + 6 y 3 + x y 2 ) = y − 8 y 2 ( 2 x + y ) 3 − 6 x y 2 y 2 {\displaystyle y'({\frac {4y^{2}(2x+y)^{3}+6y^{3}+x}{y^{2}}})={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{y^{2}}}}
使用对数求导法求 d y d x {\displaystyle {\frac {dy}{dx}}}
y ′ y = 1 x − 3 x 2 4 ( 1 − x 3 ) {\displaystyle {\frac {y'}{y}}={\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}}} y ′ = x ( 1 − x 3 4 ) ( 1 x − 3 x 2 4 ( 1 − x 3 ) ) = 1 − x 3 4 − 3 x 3 4 ( 1 − x 3 ) 3 / 4 {\displaystyle y'=x({\sqrt[{4}]{1-x^{3}}}\,)({\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}})=\mathbf {{\sqrt[{4}]{1-x^{3}}}-{\frac {3x^{3}}{4(1-x^{3})^{3/4}}}} }
y ′ y = 1 2 ( 1 x + 1 + 1 1 − x ) {\displaystyle {\frac {y'}{y}}={\frac {1}{2}}({\frac {1}{x+1}}+{\frac {1}{1-x}})}
y ′ y = 2 ln ( 2 x ) + 2 x 2 2 x = 2 ln ( 2 x ) + 2 {\displaystyle {\frac {y'}{y}}=2\ln(2x)+2x{\frac {2}{2x}}=2\ln(2x)+2}
y ′ y = 3 ln ( x 3 + 4 x ) + ( 3 x + 1 ) 3 x 2 + 4 x 3 + 4 x {\displaystyle {\frac {y'}{y}}=3\ln(x^{3}+4x)+(3x+1){\frac {3x^{2}+4}{x^{3}+4x}}}
y ′ y = − sin ( x ) ln ( 6 x ) + cos ( x ) + 1 x {\displaystyle {\frac {y'}{y}}=-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}}}
对于每个函数, f {\displaystyle f} ,(a) 确定对哪些 x {\displaystyle x} 值, f {\displaystyle f} 的切线是水平的,(b) 求过给定点的 f {\displaystyle f} 的切线方程。
a) x 2 + 2 x = 0 ⟹ x = 0 , − 2 {\displaystyle x^{2}+2x=0\implies \mathbf {x=0,-2} } b) m = 3 2 + 2 ( 3 ) = 9 + 6 = 15 {\displaystyle m=3^{2}+2(3)=9+6=15}
a) 3 x 2 − 3 = 0 ⟹ x = ± 1 {\displaystyle 3x^{2}-3=0\implies \mathbf {x=\pm 1} } b) m = 3 ( 1 ) 2 − 3 = 0 {\displaystyle m=3(1)^{2}-3=0}
a) 2 x 2 + 2 x − 12 = 0 = x 2 + x − 6 = ( x + 3 ) ( x − 2 ) ⟹ x = 2 , − 3 {\displaystyle 2x^{2}+2x-12=0=x^{2}+x-6=(x+3)(x-2)\implies \mathbf {x=2,-3} } b) m = − 12 {\displaystyle m=-12}
a) 2 − 1 2 x 3 / 2 = 0 ⟹ 2 x 3 / 2 = 1 2 ⟹ x 3 / 2 = 1 4 = 2 − 2 ⟹ x = 2 − 4 / 3 {\displaystyle 2-{\frac {1}{2x^{3/2}}}=0\implies 2x^{3/2}={\frac {1}{2}}\implies x^{3/2}={\frac {1}{4}}=2^{-2}\implies \mathbf {x=2^{-4/3}} } b) m = 2 − 1 2 ( 1 ) 3 / 2 = 2 − 1 2 = 3 2 {\displaystyle m=2-{\frac {1}{2(1)^{3/2}}}=2-{\frac {1}{2}}={\frac {3}{2}}}
a) − 3 x 2 + 4 x − 1 = 0 = ( 3 x − 1 ) ( − x + 1 ) ⟹ x = 1 , 1 3 {\displaystyle -3x^{2}+4x-1=0=(3x-1)(-x+1)\implies \mathbf {x=1,{\frac {1}{3}}} } b) m = − 3 ( 2 ) 2 + 4 ( 2 ) − 1 = − 3 ( 4 ) + 8 − 1 = − 12 + 7 = − 5 {\displaystyle m=-3(2)^{2}+4(2)-1=-3(4)+8-1=-12+7=-5}
a) 2 x 2 + 5 x + 2 = 0 = ( 2 x + 1 ) ( x + 2 ) ⟹ x = − 1 2 , − 2 {\displaystyle 2x^{2}+5x+2=0=(2x+1)(x+2)\implies \mathbf {x=-{\frac {1}{2}},-2} } / b) m = 2 ( 3 ) 2 + 5 ( 3 ) + 2 = 18 + 15 + 2 = 35 {\displaystyle m=2(3)^{2}+5(3)+2=18+15+2=35}
1 − y ′ = 1 3 ( x − y − 1 ) 2 {\displaystyle 1-y'={\frac {1}{3(x-y-1)^{2}}}} y ′ = 1 − 1 3 ( x − y − 1 ) 2 {\displaystyle y'=1-{\frac {1}{3(x-y-1)^{2}}}} m = 1 − 1 3 ( 1 − ( − 1 ) − 1 ) 2 = 1 − 1 3 ( 1 ) 2 = 2 3 {\displaystyle m=1-{\frac {1}{3(1-(-1)-1)^{2}}}=1-{\frac {1}{3(1)^{2}}}={\frac {2}{3}}} − 1 = 2 3 ( 1 ) + b ⟹ b = − 5 3 {\displaystyle -1={\frac {2}{3}}(1)+b\implies b=-{\frac {5}{3}}}
y ′ ( e x y − 2 y ) = − 2 x {\displaystyle y'(e^{xy}-2y)=-2x} y ′ = 2 x 2 y − e x y {\displaystyle y'={\frac {2x}{2y-e^{xy}}}} m = 2 ( 1 ) 2 ( 0 ) − e 1 ( 0 ) = 2 − 1 = − 2 {\displaystyle m={\frac {2(1)}{2(0)-e^{1(0)}}}={\frac {2}{-1}}=-2} 0 = − 2 ( 1 ) + b ⟹ b = 2 {\displaystyle 0=-2(1)+b\implies b=2}
归纳步骤: 假设(n-1)阶多项式的n阶导数为0。 考虑n阶多项式, f ( x ) {\displaystyle f(x)} . 我们可以写 f ( x ) = c x n + P ( x ) {\displaystyle f(x)=cx^{n}+P(x)} ,其中 P ( x ) {\displaystyle P(x)} 是(n-1)阶多项式。
g ′ ( x ) = lim h → 0 g ( x + h ) − g ( h ) h = lim h → 0 − f ( x + h ) + f ( h ) h = lim h → 0 − ( f ( x + h ) − f ( h ) ) h = − lim h → 0 ( f ( x + h ) − f ( h ) ) h = − f ′ ( x ) {\displaystyle {\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}}
如果可能,求 a {\displaystyle a} 的值,使以下每个条件都成立。如果不可能,请证明。
(b) 不存在 a ∈ R {\displaystyle a\in \mathbb {R} } 使函数可微且连续。
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