Inicio › Matemáticas 5 › Regla de la cadena (P2)

Matemáticas 5 · Tema 10

Regla de la cadena (P2)

Apuntes

La misma regla de la cadena, ahora con senos, cosenos, logaritmos y raíces:

h(x)=f(g(x))⟶h′(x)=[f′(g(x))][g′(x)]h(x) = f(g(x)) \longrightarrow h'(x) = [f'(g(x))][g'(x)]

Ejemplos

g=cos⁡3x5g=(cos⁡x)3/5g′=35(cos⁡x)35−55(−sen⁡x)g′=35(cos⁡x)−25(−sen⁡x)g′=−3sen⁡x5(cos⁡x)2/5g′=−3sen⁡x5cos⁡2x5\begin{aligned} g &= \sqrt[5]{\cos^3 x} \\ g &= (\cos x)^{3/5} \\ g' &= \frac{3}{5}(\cos x)^{\frac{3}{5}-\frac{5}{5}}(-\operatorname{sen} x) \\ g' &= \frac{3}{5}(\cos x)^{-\frac{2}{5}}(-\operatorname{sen} x) \\ g' &= -\frac{3\operatorname{sen} x}{5(\cos x)^{2/5}} \\ \mathbf{g'} &= \mathbf{-\frac{3\operatorname{sen} x}{5\sqrt[5]{\cos^2 x}}} \end{aligned}
h=cos⁡(3x−5x2)h′=−sen⁡(3x−5x2)[3−10x]h′=−3sen⁡(3x−5x2)+10xsen⁡(3x−5x2)\begin{aligned} h &= \cos(3x - 5x^2) \\ h' &= -\operatorname{sen}(3x - 5x^2)[3 - 10x] \\ \mathbf{h'} &= \mathbf{-3\operatorname{sen}(3x - 5x^2) + 10x\operatorname{sen}(3x - 5x^2)} \end{aligned}
g(x)=ln⁡(5x2−5)g′(x)=[15x2−5](10x)g′(x)=10x5x2−5\begin{aligned} g(x) &= \ln(5x^2 - 5) \\ g'(x) &= \left[\frac{1}{5x^2 - 5}\right](10x) \\ \mathbf{g'(x)} &= \mathbf{\frac{10x}{5x^2 - 5}} \end{aligned}
f=5sen⁡3xf=5(sen⁡x)3f′=15sen⁡2x⋅cos⁡x\begin{aligned} f &= 5\operatorname{sen}^3 x \\ f &= 5(\operatorname{sen} x)^3 \\ \mathbf{f'} &= \mathbf{15\operatorname{sen}^2 x \cdot \cos x} \end{aligned}

Ejercicios

Bloque 1
f=cos⁡7x3f = \sqrt[3]{\cos^7 x}
f=sen⁡3xf = \sqrt{\operatorname{sen}^3 x}
f=sen⁡5x4f = \sqrt[4]{\operatorname{sen}^5 x}
Bloque 2
f=cos⁡(7x−5x2)f = \cos(7x - 5x^2)
f=sen⁡(6x−4x2)f = \operatorname{sen}(6x - 4x^2)
f=cos⁡(2x3−9x)f = \cos(2x^3 - 9x)
Bloque 3
f=ln⁡(3x2−4)f = \ln(3x^2 - 4)
f=ln⁡(2x2−5x)f = \ln(2x^2 - 5x)
f=ln⁡(4x3+7x)f = \ln(4x^3 + 7x)
Bloque 4
f=−5sen⁡5xf = -5\operatorname{sen}^5 x
f=4cos⁡3xf = 4\cos^3 x
f=3cos⁡6xf = 3\cos^6 x

Respuestas

▼
Bloque 1
f=cos⁡7x3f=(cos⁡x)7/3f′=73(cos⁡x)73−33(−sen⁡x)f′=73(cos⁡x)4/3(−sen⁡x)f′=−7sen⁡x⋅cos⁡4x33\begin{aligned} f &= \sqrt[3]{\cos^7 x} \\ f &= (\cos x)^{7/3} \\ f' &= \frac{7}{3}(\cos x)^{\frac{7}{3} - \frac{3}{3}}(-\operatorname{sen} x) \\ f' &= \frac{7}{3}(\cos x)^{4/3}(-\operatorname{sen} x) \\ \mathbf{f'} &= \mathbf{-\frac{7\operatorname{sen} x \cdot \sqrt[3]{\cos^4 x}}{3}} \end{aligned}
f=sen⁡3xf=(sen⁡x)3/2f′=32(sen⁡x)32−22(cos⁡x)f′=32(sen⁡x)1/2(cos⁡x)f′=3cos⁡xsen⁡x2\begin{aligned} f &= \sqrt{\operatorname{sen}^3 x} \\ f &= (\operatorname{sen} x)^{3/2} \\ f' &= \frac{3}{2}(\operatorname{sen} x)^{\frac{3}{2} - \frac{2}{2}}(\cos x) \\ f' &= \frac{3}{2}(\operatorname{sen} x)^{1/2}(\cos x) \\ \mathbf{f'} &= \mathbf{\frac{3\cos x \sqrt{\operatorname{sen} x}}{2}} \end{aligned}
f=sen⁡5x4f=(sen⁡x)5/4f′=54(sen⁡x)54−44(cos⁡x)f′=54(sen⁡x)1/4(cos⁡x)f′=5cos⁡xsen⁡x44\begin{aligned} f &= \sqrt[4]{\operatorname{sen}^5 x} \\ f &= (\operatorname{sen} x)^{5/4} \\ f' &= \frac{5}{4}(\operatorname{sen} x)^{\frac{5}{4} - \frac{4}{4}}(\cos x) \\ f' &= \frac{5}{4}(\operatorname{sen} x)^{1/4}(\cos x) \\ \mathbf{f'} &= \mathbf{\frac{5\cos x \sqrt[4]{\operatorname{sen} x}}{4}} \end{aligned}
Bloque 2
f=cos⁡(7x−5x2)f′=−sen⁡(7x−5x2)[7−10x]f′=−7sen⁡(7x−5x2)+10xsen⁡(7x−5x2)\begin{aligned} f &= \cos(7x - 5x^2) \\ f' &= -\operatorname{sen}(7x - 5x^2)[7 - 10x] \\ \mathbf{f'} &= \mathbf{-7\operatorname{sen}(7x - 5x^2) + 10x\operatorname{sen}(7x - 5x^2)} \end{aligned}
f=sen⁡(6x−4x2)f′=cos⁡(6x−4x2)[6−8x]f′=6cos⁡(6x−4x2)−8xcos⁡(6x−4x2)\begin{aligned} f &= \operatorname{sen}(6x - 4x^2) \\ f' &= \cos(6x - 4x^2)[6 - 8x] \\ \mathbf{f'} &= \mathbf{6\cos(6x - 4x^2) - 8x\cos(6x - 4x^2)} \end{aligned}
f=cos⁡(2x3−9x)f′=−sen⁡(2x3−9x)[6x2−9]f′=−6x2sen⁡(2x3−9x)+9sen⁡(2x3−9x)\begin{aligned} f &= \cos(2x^3 - 9x) \\ f' &= -\operatorname{sen}(2x^3 - 9x)[6x^2 - 9] \\ \mathbf{f'} &= \mathbf{-6x^2\operatorname{sen}(2x^3 - 9x) + 9\operatorname{sen}(2x^3 - 9x)} \end{aligned}
Bloque 3
f=ln⁡(3x2−4)f′=[13x2−4](6x)f′=6x3x2−4\begin{aligned} f &= \ln(3x^2 - 4) \\ f' &= \left[\frac{1}{3x^2 - 4}\right](6x) \\ \mathbf{f'} &= \mathbf{\frac{6x}{3x^2 - 4}} \end{aligned}
f=ln⁡(2x2−5x)f′=[12x2−5x](4x−5)f′=4x−52x2−5x\begin{aligned} f &= \ln(2x^2 - 5x) \\ f' &= \left[\frac{1}{2x^2 - 5x}\right](4x - 5) \\ \mathbf{f'} &= \mathbf{\frac{4x - 5}{2x^2 - 5x}} \end{aligned}
f=ln⁡(4x3+7x)f′=[14x3+7x](12x2+7)f′=12x2+74x3+7x\begin{aligned} f &= \ln(4x^3 + 7x) \\ f' &= \left[\frac{1}{4x^3 + 7x}\right](12x^2 + 7) \\ \mathbf{f'} &= \mathbf{\frac{12x^2 + 7}{4x^3 + 7x}} \end{aligned}
Bloque 4
f=−5sen⁡5xf=−5(sen⁡x)5f′=−25(sen⁡x)4⋅cos⁡xf′=−25sen⁡4x⋅cos⁡x\begin{aligned} f &= -5\operatorname{sen}^5 x \\ f &= -5(\operatorname{sen} x)^5 \\ f' &= -25(\operatorname{sen} x)^4 \cdot \cos x \\ \mathbf{f'} &= \mathbf{-25\operatorname{sen}^4 x \cdot \cos x} \end{aligned}
f=4cos⁡3xf=4(cos⁡x)3f′=12(cos⁡x)2(−sen⁡x)f′=−12cos⁡2x⋅sen⁡x\begin{aligned} f &= 4\cos^3 x \\ f &= 4(\cos x)^3 \\ f' &= 12(\cos x)^2(-\operatorname{sen} x) \\ \mathbf{f'} &= \mathbf{-12\cos^2 x \cdot \operatorname{sen} x} \end{aligned}
f=3cos⁡6xf=3(cos⁡x)6f′=18(cos⁡x)5(−sen⁡x)f′=−18cos⁡5x⋅sen⁡x\begin{aligned} f &= 3\cos^6 x \\ f &= 3(\cos x)^6 \\ f' &= 18(\cos x)^5(-\operatorname{sen} x) \\ \mathbf{f'} &= \mathbf{-18\cos^5 x \cdot \operatorname{sen} x} \end{aligned}

Ejercicios extra 1

▼
Bloque A
y=cos⁡3x4y = \sqrt[4]{\cos^3 x}
y=cos⁡5xy = \sqrt{\cos^5 x}
y=sen⁡2x3y = \sqrt[3]{\operatorname{sen}^2 x}
Bloque B
y=cos⁡(8x−3x2)y = \cos(8x - 3x^2)
y=sen⁡(7x−2x3)y = \operatorname{sen}(7x - 2x^3)
y=cos⁡(4x2−9x)y = \cos(4x^2 - 9x)
Bloque C
y=ln⁡(5x2−2x)y = \ln(5x^2 - 2x)
y=ln⁡(x3−4x)y = \ln(x^3 - 4x)
y=ln⁡(3x3−6x)y = \ln(3x^3 - 6x)
Bloque D
y=−3sen⁡4xy = -3\operatorname{sen}^4 x
y=7sen⁡3xy = 7\operatorname{sen}^3 x
y=−2cos⁡5xy = -2\cos^5 x

Ejercicios extra 2

▼
Bloque E
y=sen⁡4x5y = \sqrt[5]{\operatorname{sen}^4 x}
y=sen⁡(5x3−2x2)y = \operatorname{sen}(5x^3 - 2x^2)
y=ln⁡(6x2−3)y = \ln(6x^2 - 3)
y=6cos⁡4xy = 6\cos^4 x
← Regla de la cadena (P1) Recta tangente →