1)
lim
h
→
0
(
2
+
h
)
2
−
4
h
{\displaystyle \lim _{h\to 0}{\frac {(2+h)^{2}-4}{h}}}
lim
h
→
0
4
+
4
h
+
h
2
−
4
h
{\displaystyle \lim _{h\to 0}{\frac {4+4h+h^{2}-4}{h}}}
lim
h
→
0
4
h
+
h
2
h
{\displaystyle \lim _{h\to 0}{\frac {4h+h^{2}}{h}}}
lim
h
→
0
h
(
4
+
h
)
h
{\displaystyle \lim _{h\to 0}{\frac {h(4+h)}{h}}}
lim
h
→
0
4
+
h
{\displaystyle \lim _{h\to 0}\ {4+h}}
4
+
0
{\displaystyle \ {4+0}}
4
{\displaystyle \ {4}}
Jorgemontes
2)Derivar:
F
(
x
)
=
(
x
+
1
)
2
x
−
1
{\displaystyle \ F(x)={\frac {(x+1)^{2}}{x-1}}}
lim
h
→
0
(
(
(
x
+
h
)
+
1
)
2
)
(
x
+
h
)
−
1
−
(
x
+
1
)
2
x
−
1
h
{\displaystyle \lim _{h\to 0}{\frac {{\frac {(((x+h)+1)^{2})}{(x+h)-1}}-{\frac {(x+1)^{2}}{x-1}}}{h}}}
lim
h
→
0
h
x
2
+
h
2
x
−
3
h
−
2
h
x
(
x
+
h
−
1
)
(
x
−
1
)
h
{\displaystyle \lim _{h\to 0}{\frac {\frac {hx^{2}+h^{2}x-3h-2hx}{(x+h-1)(x-1)}}{h}}}
lim
h
→
0
(
h
)
(
x
2
+
h
x
−
3
−
2
x
)
(
h
)
(
(
x
+
h
−
1
)
(
x
−
1
)
)
{\displaystyle \lim _{h\to 0}{\frac {(h)(x^{2}+hx-3-2x)}{(h)((x+h-1)(x-1))}}}
lim
h
→
0
x
2
+
h
x
−
3
−
2
x
(
x
+
h
−
1
)
(
x
−
1
)
{\displaystyle \lim _{h\to 0}{\frac {x^{2}+hx-3-2x}{(x+h-1)(x-1)}}}
x
2
−
2
x
−
3
(
x
−
1
)
2
{\displaystyle {\frac {x^{2}-2x-3}{(x-1)^{2}}}}
Jorgemontes 3)Hallar la derivada de:
y=x
y
′
=
1
{\displaystyle \ y'=1}
l
i
m
x
→
0
f
(
x
+
h
)
−
F
(
x
)
h
{\displaystyle \ lim_{x\to 0}{\frac {f(x+h)-F(x)}{h}}}
l
i
m
x
→
0
x
+
h
−
x
h
{\displaystyle \ lim_{x\to 0}{\frac {x+h-x}{h}}}
l
i
m
x
→
0
h
h
{\displaystyle \ lim_{x\to 0}{\frac {h}{h}}}
1
{\displaystyle \ 1}
by: jorge montes and Zorraidorsito Jorgemontes
4)Derive:
6
x
2
+
1
{\displaystyle {\frac {6}{x^{2}+1}}}
lim
h
→
0
6
(
x
+
h
)
2
+
1
−
6
x
2
+
1
h
{\displaystyle \lim _{h\to 0}{\frac {{\frac {6}{(x+h)^{2}+1}}-{\frac {6}{x^{2}+1}}}{h}}}
lim
h
→
0
6
x
2
+
6
−
6
x
2
−
12
h
x
−
6
h
2
−
6
(
(
x
+
h
)
2
+
1
)
(
x
2
+
1
)
h
{\displaystyle \lim _{h\to 0}{\frac {\frac {6x^{2}+6-6x^{2}-12hx-6h^{2}-6}{((x+h)^{2}+1)(x^{2}+1)}}{h}}}
lim
h
→
0
−
6
h
2
−
12
h
x
(
(
x
+
h
)
2
+
1
)
(
x
2
+
1
)
h
{\displaystyle \lim _{h\to 0}{\frac {\frac {-6h^{2}-12hx}{((x+h)^{2}+1)(x^{2}+1)}}{h}}}
lim
h
→
0
−
6
h
2
−
12
h
x
(
h
)
(
(
(
x
+
h
)
2
+
1
)
(
x
2
+
1
)
)
{\displaystyle \lim _{h\to 0}{\frac {-6h^{2}-12hx}{(h)(((x+h)^{2}+1)(x^{2}+1))}}}
lim
h
→
0
(
h
)
(
−
6
h
−
12
x
)
(
h
)
(
(
(
x
+
h
)
2
+
1
)
(
x
2
+
1
)
)
{\displaystyle \lim _{h\to 0}{\frac {(h)(-6h-12x)}{(h)(((x+h)^{2}+1)(x^{2}+1))}}}
lim
h
→
0
−
6
h
−
12
x
(
(
x
+
h
)
2
+
1
)
(
x
2
+
1
)
{\displaystyle \lim _{h\to 0}{\frac {-6h-12x}{((x+h)^{2}+1)(x^{2}+1)}}}
−
6
(
0
)
−
12
x
(
(
x
+
(
0
)
)
2
+
1
)
(
x
2
+
1
)
{\displaystyle {\frac {-6(0)-12x}{((x+(0))^{2}+1)(x^{2}+1)}}}
−
12
x
(
x
2
+
1
)
2
{\displaystyle {\frac {-12x}{(x^{2}+1)^{2}}}}
Jorgemontes 5)Derive por implícita:
y
2
−
x
2
=
1
{\displaystyle \ {y^{2}-x^{2}=1}}
d
d
x
y
2
−
d
d
x
x
2
=
d
d
x
1
{\displaystyle \ {{\frac {d}{dx}}y^{2}-{\frac {d}{dx}}x^{2}={\frac {d}{dx}}1}}
d
d
x
2
y
−
d
d
x
2
x
=
d
d
x
0
{\displaystyle \ {{\frac {d}{dx}}2y-{\frac {d}{dx}}2x={\frac {d}{dx}}0}}
d
d
x
2
y
−
d
d
x
2
x
=
0
{\displaystyle \ {{\frac {d}{dx}}2y-{\frac {d}{dx}}2x=0}}
d
d
x
(
2
y
−
2
x
)
=
0
{\displaystyle \ {{\frac {d}{dx}}(2y-2x)=0}}
d
d
x
(
2
y
)
=
2
x
{\displaystyle \ {{\frac {d}{dx}}(2y)=2x}}
d
d
x
=
2
x
2
y
{\displaystyle \ {{\frac {d}{dx}}={\frac {2x}{2y}}}}
d
d
x
=
x
y
{\displaystyle \ {{\frac {d}{dx}}={\frac {x}{y}}}}
6)calcular
f
,
(
0
)
,
f
,
(
1
2
)
,
f
,
(
1
)
,
f
,
(
−
10
)
{\displaystyle \ f^{,}(0),f^{,}({\frac {1}{2}}),f^{,}(1),f^{,}(-10)}
f
(
x
)
=
2
+
x
−
x
2
{\displaystyle \ {f(x)=2+x-x^{2}}}
f
,
(
x
)
=
1
−
2
x
{\displaystyle \ {f^{,}(x)=1-2x}}
f
,
(
0
)
=
1
{\displaystyle \ {f^{,}(0)=1}}
f
,
(
1
2
)
=
0
{\displaystyle \ {f^{,}({\frac {1}{2}})=0}}
f
,
(
1
)
=
−
1
{\displaystyle \ {f^{,}(1)=-1}}
f
,
(
−
10
)
=
−
19
{\displaystyle \ {f^{,}(-10)=-19}}
Encuentre la ecuación de la recta tangente en el punto propuesto
1)
x
3
y
+
y
3
x
=
30
{\displaystyle \ {x^{3}y+y^{3}x=30}}
![{\displaystyle {\frac {1}{\sqrt[{2}]{3x}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7baf93bbda6139f64ecb83069db088a452516a06)
![{\displaystyle \lim _{h\to 0}{\frac {{\frac {1}{\sqrt[{2}]{3x+3h}}}-{\frac {1}{\sqrt[{2}]{3x}}}}{h}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33c480795fc1b0328e3b9dfd940b49f1c736cb12)
