Strona 1 z 1
Kilka pochodnych..
: 30 gru 2008, o 19:29
autor: Zenek1
1.y=\(\displaystyle{ x^{2}}\)\(\displaystyle{ e^{2x}}\)-x\(\displaystyle{ e^{2x}}\)-\(\displaystyle{ e^{2}}\)
2.y=\(\displaystyle{ 2^{x^{2}{+1}}}\)
3.y=\(\displaystyle{ x^{2}}\)\(\displaystyle{ \sqrt{1+\sqrt{x}}}\)
Kilka pochodnych..
: 30 gru 2008, o 19:42
autor: soku11
1.
\(\displaystyle{ y=x^2e^{2x}-xe^{2x}-e^2\\
y'=(x^2e^{2x}-xe^{2x}-e^2)'=
(x^2e^{2x})'-(xe^{2x})'-(e^2)'=
(x^2)'e^{2x}+x^2(e^{2x})'-[(x)'e^{2x}+x(e^{2x})']-0=
2xe^{2x}+2x^2e^{2x}-[e^{2x}+2xe^{2x}]=
2xe^{2x}+2x^2e^{2x}-e^{2x}-2xe^{2x}=
2x^2e^{2x}-e^{2x}}\)
2.
\(\displaystyle{ y=2^{x^2+1}\\
y'=2^{x^2+1}\cdot \ln 2\cdot (x^2+1)'=
2x2^{x^2+1}\ln 2}\)
3.
\(\displaystyle{ y=x^2\cdot\sqrt{1+\sqrt{x}}\\
y'=(x^2\cdot\sqrt{1+\sqrt{x}})'=
(x^2)'\cdot\sqrt{1+\sqrt{x}}+x^2\cdot(\sqrt{1+\sqrt{x}})'=
2x\cdot\sqrt{1+\sqrt{x}}+x^2\cdot\frac{(1+\sqrt{x})'}{2\sqrt{1+\sqrt{x}}}=
2x\cdot\sqrt{1+\sqrt{x}}+x^2\cdot\frac{1}{4\sqrt{x}\sqrt{1+\sqrt{x}}}}\)
Pozdrawiam.