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Pochodna funkcji jednej zmiennej - 20.03

1

Korzystając ze wzorów bezpośrednich i praw działań na pochodnych obliczymy pochodne funkcji:

  1. f(x)=4x^5-3x^2+x\sqrt[3]x-\ln 3,
  2. f(x)=6x^3\cdot \operatorname{arcsin} x,
  3. f(x)=\frac{3x^4-5x+1}{x^2},
  4. f(x)=\frac1x\cdot\cos x,
  5. f(x)=\frac{1+\ln x}{x^4+\pi}.

  1. f^\prime(x)=\left(4x^5-3x^2+x\sqrt[3]x-\ln 3\right)^\prime=4\left(x^5\right)^\prime-3\left(x^2\right)^\prime+\left(x^{\frac43}\right)^\prime-\left(\ln 3\right)^\prime=

    =4\cdot5x^4-3\cdot2x+\frac43x^{\frac13}-0=20x^4-6x+\frac43x^{\frac13}.

  2. f^\prime(x)=\left(6x^3\cdot \operatorname{arcsin} x\right)^\prime=6\left(x^3\right)^\prime\cdot \operatorname{arcsin} x+6x^3\cdot\left(\operatorname{arcsin} x\right)^\prime=

    =18x^2\cdot \operatorname{arcsin}x+6x^3 \cdot \frac{1}{\sqrt{1-x^2}}=18x^2\cdot \operatorname{arcsin}x+\frac{6x^3}{\sqrt{1-x^2}}.

  3. f^\prime(x)=\left(\frac{3x^4-5x+1}{x^2}\right)^\prime=\frac{\left(3x^4-5x+1\right)^\prime\cdot x^2-\left(3x^4-5x+1\right) \cdot\left(x^2\right)^\prime}{\left(x^2\right)^2}=

    \frac{\left(12x^3-5\right)\cdot x^2-\left(3x^4-5x+1\right)\cdot 2x}{x^4}= \frac{12x^5-5x^2-6x^5+10x^2-2x}{x^4}=

    =\frac{6x^5+5x^2-2x}{x^4}=\frac{x\left(6x^4+5x-2\right)}{x^4}=\frac{6x^4+5x-2}{x^3}.

  4. f^\prime(x)=\left(\frac1x\cdot\cos x\right)^\prime=\left(\frac{\cos x}{x}\right)^\prime=\frac{\left(\cos x\right)^\prime\cdot x-\cos x\cdot \left(x\right)^\prime}{x^2}=

    =\frac{-\sin x\cdot x-\cos x\cdot 1}{x^2}=-\frac{x\sin x+\cos x}{x^2}.

  5. f^\prime(x)=\left(\frac{1+\ln x}{x^4+\pi}\right)^\prime=\frac{\left(1+\ln x\right)^\prime\cdot \left(x^4+\pi\right)-\left(1+\ln x\right)\cdot\left(x^4+\pi\right)^\prime}{\left(x^4+\pi\right)^2}=

    =\frac{\left(0+\frac 1x\right)\cdot \left(x^4+\pi\right)-\left(1+\ln x\right)\cdot \left(4x^3+0\right)}{\left(x^4+\pi\right)^2}=\frac{x^3+\frac{\pi}x-4x^3-4x^3\ln x}{\left(x^4+\pi\right)^2}=

    =\frac{-3x^3+\frac{\pi}x-4x^3\ln x}{\left(x^4+\pi\right)^2} =\frac{-x^3\left(3+4\ln x\right)+\frac{\pi}x}{\left(x^4+\pi\right)^2}.

2

Korzystając ze wzorów bezpośrednich i praw działań na pochodnych oraz twierdzenia o pochodnej funkcji złożonej obliczymy pochodne funkcji:

  1. f(x)=\ln 8x,
  2. f(x)=\operatorname{arcctg}(x^6),
  3. f(x)=\text{tg}^{4} x,
  4. f(x)=e^{-2x}\cdot \operatorname{arcccos}(3x),
  5. f(x)=\frac{x^3\cos 3x}{1+x^2},
  6. f(x)=\sin^5(\cos 7x).

  1. f^\prime(x)=\left(\ln 8x\right)^\prime=\frac1{8x}\cdot \left(8x\right)^\prime=\frac1{8x}\cdot 8=\frac{1}{x}.

  2. f^\prime(x)=\left(\operatorname{arcctg}(x^6)\right)^\prime=-\frac {1}{1+\left(x^6\right)^2}\cdot \left(x^6\right)^\prime=-\frac {1}{1+x^{12}}\cdot 6x^5 =-\frac {6x^5}{1+x^{12}}.

  3. f^\prime(x)=\left(\text{tg}^4 x\right)^\prime=\left(\left(\text{tg} x\right)^4\right)^\prime=4\left(\text{tg} x\right)^3\cdot \left(\text{tg} x\right)^\prime=4\text{tg}^3 x\cdot \frac1{\cos^2 x}=\frac{4\text{tg}^3 x}{\cos^2 x}.

  4. f^\prime(x)=\left(e^{-2x}\cdot \operatorname{arcccos}(3x)\right)^\prime=\left(e^{-2x}\right)^\prime\cdot \operatorname{arcccos}(3x)+e^{-2x}\cdot \left(\operatorname{arcccos}(3x)\right)^\prime=

    =e^{-2x}\cdot\left(-2x\right)^\prime\cdot\operatorname{arcccos}(3x)+e^{-2x}\cdot\left(-\frac{1}{\sqrt{1-\left(3x\right)^2}}\right)\cdot\left(3x\right)^\prime=

    =e^{-2x}\cdot\left(-2\right)\cdot\operatorname{arcccos}(3x)+e^{-2x}\cdot\left(-\frac{1}{\sqrt{1-9x^2}}\right)\cdot3= =-2e^{-2x}\cdot\operatorname{arcccos}(3x)-\frac{3e^{-2x}}{\sqrt{1-9x^2}}.

  5. f^\prime(x)=\left(\frac{x^3\cos 3x}{1+x^2}\right)^\prime=\frac{\left(x^3\cos 3x\right)^\prime\cdot\left(1+x^2\right)-\left(x^3\cos 3x\right)\cdot\left(1+x^2\right)^\prime}{\left(1+x^2\right)^2}=

    =\frac{\left(\left(x^3\right)^\prime\cdot\cos3x+x^3\cdot\left(\cos3x\right)^\prime\right) \cdot\left(1+x^2\right)-x^3\cos 3x\cdot\left(0+2x\right)}{\left(1+x^2\right)^2}=

    =\frac{\left(3x^2\cos3x+x^3\cdot\left(-\sin3x\right)\cdot\left(3x\right)^\prime\right) \cdot\left(1+x^2\right)-2x^4\cos 3x}{\left(1+x^2\right)^2}=

    =\frac{\left(3x^2\cos3x-3x^3\cdot\sin3x\right)\cdot\left(1+x^2\right)-2x^4\cos 3x}{\left(1+x^2\right)^2}.

  6. f^\prime(x)=\left(\sin^5(\cos7x)\right)^\prime=\left(\left(\sin\left(\cos7x\right)\right)^5\right)^\prime=

    =5\left(\sin\left(\cos7x\right)\right)^4\cdot \left(\sin\left(\cos7x\right)\right)^\prime=5\sin^4\left(\cos7x\right)\cdot \left(\cos\left(\cos7x\right)\right)\cdot\left(\cos7x\right)^\prime=

    =5\sin^4\left(\cos7x\right)\cdot \left(\cos\left(\cos7x\right)\right)\cdot\left(-\sin7x\right)\cdot\left(7x\right)^\prime =

    =-5\sin^4\left(\cos7x\right)\cdot \left(\cos\left(\cos7x\right)\right)\cdot\sin7x\cdot7=-35\sin^4\left(\cos7x\right)\cdot \left(\cos\left(\cos7x\right)\right)\cdot\sin7x.

3

Obliczymy pochodne funkcji:

  1. f(x)=(x)^{5x},
  2. f(x)=\log_{x^2}(e^x).

  1. f^\prime(x)=\left((x)^{5x}\right)^\prime=\left(e^{5x\cdot\ln x}\right)^\prime=e^{5x\cdot\ln x}\cdot \left(5x\cdot\ln x\right)^\prime=5e^{5x\cdot\ln x}\cdot \left(\left(x\right)^\prime\cdot\ln x+x\cdot\left(\ln x\right)^\prime\right)=

    =5e^{5x\cdot\ln x}\cdot \left(1\cdot\ln x+x\cdot\frac1x\right)=5e^{5x\cdot\ln x}\cdot \left(\ln x+1\right)=5(x)^{5x}\cdot \left(\ln x+1\right).

  2. f^\prime(x)=\left(\log_{x^2}(e^x)\right)^\prime=\left(\frac{\ln e^x}{\ln x^2}\right)^\prime=\frac{\left(x\right)^\prime\cdot\ln x^2-x\cdot\left(\ln x^2\right)^\prime}{\left(\ln x^2\right)^2}=

    =\frac{1\cdot\ln x^2-x\cdot\frac1{x^2}\cdot 2x}{\left(\ln x^2\right)^2}=\frac{\ln x^2-2}{\ln^2 x^2}.

4

Obliczymy pochodną funkcji f(x)=\frac{\ln (x^3+2x+1)}{e^{x^2}} w punkcie x_0=1.

f^\prime(x)=\left(\frac{\ln (x^3+2x+1)}{e^{x^2}}\right)^\prime=\frac{\left(\ln (x^3+2x+1)\right)^\prime\cdot e^{x^2}-\ln (x^3+2x+1)\cdot \left(e^{x^2}\right)^\prime}{\left(e^{x^2}\right)^2}=

=\frac{\frac1{x^3+2x+1}\cdot\left(x^3+2x+1\right)^\prime\cdot e^{x^2}-\ln (x^3+2x+1)\cdot e^{x^2}\cdot\left(x^2\right)^\prime }{e^{2x^2}}=

=\frac{\frac1{x^3+2x+1}\cdot\left(3x^2+2\right)\cdot e^{x^2}-\ln (x^3+2x+1)\cdot 2xe^{x^2} }{e^{2x^2}},

f^\prime(1)=\frac{\frac1{1^3+2\cdot 1+1}\cdot\left((3\cdot 1^2+2)\right)\cdot e^{1^2}-\ln (1^3+2\cdot 1+1)\cdot 2\cdot1\cdot e^{1^2} }{e^{2\cdot1^2}}= =\frac{\frac{5e}{4}-2e\ln4}{e^2}=\frac{e(5-8\ln4)}{4e^2}=\frac{5-8\ln4}{4e}.