Temat:
MatematykaAutor:
eliseoUtworzono:
1 rok temu[tex]\left(e^{3x}\right)'}=e^{3x}\cdot\left(3x\right)'=e^{3x}\cdot3=3e^{3x}[/tex]
b)[tex]\left(e^{\sin x}\right)'}=e^{\sin x}\cdot\left(\sin x\right)'=e^{\sin x}\cos x[/tex]
c)[tex]\left(\sqrt[5]{x^2}\right)'}=\left(x^\frac25}\right)'}=\frac25\big x^{\frac25-1}=\frac25\big x^{-\frac35}=\dfrac2{5x^\frac35}=\dfrac2{5\sqrt[5]{x^3}}[/tex]
d)[tex]\left(\dfrac2{\sqrt[3]{x^2}}\right)'}=\left(2\big x^{-\frac23}}\right)'}=2\cdot(-\frac23)\big x^{-\frac23-1}=-\frac43\big x^{-\frac53}=-\dfrac4{3\sqrt[3]{x^5}}[/tex]
e)[tex]\left(\dfrac{3x^2}{7x^5-x+2} \right)'=\dfrac{(3x^2)'\cdot(7x^5-x+2)-(7x^5-x+2)'\cdot3x^2}{(7x^5-x+2)^2}=\\\\\\=\dfrac{6x\cdot(7x^5-x+2)-(35x^4-1)\cdot3x^2}{(7x^5-x+2)^2}=\dfrac{3x\cdot(14x^5-2x+4-35x^5+x)}{(7x^5-x+2)^2}=\\\\\\=\dfrac{3x(-21x^5-x+4)}{(7x^5-x+2)^2}[/tex]
f)[tex]\left(\big(3x+1\big)^7\right)'=7\cdot\big(3x+1\big)^6\left(3x+1\right)'=7\cdot\big(3x+1\big)^6\cdot3=21\big(3x+1\big)^6[/tex]
g)[tex]\left(\big(4x^2-5x+13\big)^5\right)'=5\cdot\big(4x^2-5x+13\big)^4\left(4x^2-5x+13\right)' = \\\\=5\cdot\big(4x^2-5x+13\big)^4\cdot(8x-5)=5(8x-5)\big(4x^2-5x+13\big)^4[/tex]
Autor:
alex0j3w
Oceń odpowiedź:
5