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vezi atasament , de ce nu ma lasa doar sa atasez?
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[tex]\displaystyle \mathtt{a)~f(x)=2x^2+5x}\\ \\ \mathtt{f'(x)=\left(2x^2+5x\right)'=\left(2x^2\right)'+(5x)'=2\left(x^2\right)'+5x'=2 \cdot 2x+5\cdot 1=}\\ \\ \mathtt{=4x+5}\\ \\ \mathtt{f''(x)=(f'(x))'=(4x+5)'=(4x)'+5'=4x'+0=4 \cdot 1=4}[/tex]

[tex]\displaystyle \mathtt{b)~f(x)=x^3-4x}\\ \\ \mathtt{f'(x)=\left(x^3-4x\right)'=\left(x^3\right)'-(4x)'=3x^2-4x'=3x^2-4\cdot1=3x^2-4}\\ \\ \mathtt{f''(x)=(f'(x))'=\left(3x^2-4\right)'=\left(3x^2\right)'-4'=3\left(x^2\right)'-0=3 \cdot 2x=6x}[/tex]

[tex]\displaystyle \mathtt{c)~f(x)=e^x+x}\\ \\ \mathtt{f'(x)=\left(e^x+x\right)'=\left(e^x\right)'+x'=e^x+1}\\ \\ \mathtt{f''(x)=(f'(x))'=\left(e^x+1\right)'=\left(e^x)'+1'=e^x+0=e^xx}[/tex]

[tex]\displaystyle \mathtt{d)~f(x)=x+ln~x}\\ \\ \mathtt{f'(x)=(x+ln~x)'=x'+(ln~x)'=1+ \frac{1}{x}}\\ \\ \mathtt{f''(x)=\left(1+ \frac{1}{x}\right)'=1'+\left( \frac{1}{x}\right)'=0+\left(x^{-1}\right)'=(-1)\cdot x^{-1-1}=}\\ \\ \mathtt{=-x^{-2}=- \frac{1}{x^2} }[/tex]

[tex]\displaystyle \mathtt{e)~f(x)=xln~x}\\ \\ \mathtt{f'(x)=(xln~x)'=x' \cdot ln~x+x \cdot(ln~x)'=1 \cdot ln~x+x \cdot \frac{1}{x}=ln~x+1 }\\ \\ \mathtt{f''(x)=(f'(x))'=(ln~x+1)'=(ln~x)'+1'= \frac{1}{x}+0= \frac{1}{x} }[/tex]

[tex]\displaystyle \mathtt{f)~f(x)=x^2e^x}\\ \\ \mathtt{f'(x)=\left(x^2e^x\right)'=\left(x^2\right)'\cdot e^x+x^2 \cdot \left(e^x\right)'=2xe^x+x^2e^x}\\\\\mathtt{f''(x)=(f'(x))'=\left(2xe^x+x^2e^x\right)'=\left(2xe^x\right)'+\left(x^2e^x\right)'=}\\ \\ \mathtt{=2\left(xe^x\right)'+\left(x^2e^x\right)'=2\left(x' \cdot e^x+x \cdot \left(e^x\right)'\right)+\left(x^2\right)'\cdot e^x+x^2\cdot\left(e^x\right)'=}[/tex]

[tex]\displaystyle \mathtt{=2\left(1 \cdot e^x+x e^x\right)+2xe^x+x^2e^x=2\left(e^x+xe^x\right)+2xe^x+x^2e^x=}\\ \\ \mathtt{=2e^x+2xe^x+2xe^x+x^2e^x=2e^x+4xe^x+x^2e^x} [/tex]

[tex]\displaystyle \mathtt{g)~f(x)=x^2ln~x}\\ \\ \mathtt{f'(x)=\left(x^2ln~x\right)'=\left(x^2\right)' \cdot ln~x+x^2 \cdot (ln~x)'=2xln~x+x^2 \cdot \frac{1}{x}= }\\ \\ \mathtt{=2xln~x+x}\\ \\ \mathtt{f''(x)=(f'(x))'=(2xln~x+x)'=(2xln~x)'+x'=}\\ \\ \mathtt{=(2x)'\cdot ln~x+2x \cdot (ln~x)'+1=2x' \cdot ln~x+2x \cdot \frac{1}{x}+1= }\\ \\ \mathtt{=2 \cdot 1 \cdot ln~x+2+1=2ln~x+3}[/tex]

[tex]\displaystyle \mathtt{h)~f(x)=sin^2x}\\ \\ \mathtt{f'(x)=\left(sin^2x\right)'=(sin~x\cdot sin~x)'=(sin~x)' \cdot sin~x+sin~x\cdot(sin~x)'}\\ \\ \mathtt{=cos~x \cdot sin~x+sin~x \cdot cos~x=2cos~x \cdot sin~x}\\ \\ \mathtt{f''(x)=(f'(x))'=(2cos~x \cdot sin~x)'=2(cos~x \cdot sin~x)'=}\\ \\ \mathtt{=2((cos~x)' \cdot sin~x+cos~x \cdot(sin~x)')=}\\ \\ \mathtt{=2(-sin~x\cdot sin~x+cos~x \cdot cos~x)=2(-sin^2x+cos^2x)=}\\ \\ \mathtt{=-2sin^2x+2cos^2x}[/tex]

[tex]\displaystyle \mathtt{i)~f(x)=cos^3x}\\ \\ \mathtt{f'(x)=\left(cos^3x\right)'=\left(cos^2x\cdot cos~x\right)'=}\\ \\ \mathtt{=\left(cos^2x\right)' \cdot cos~x+cos^2x \cdot (cos~x)'=}\\ \\ \mathtt{=(cos~x\cdot cos~x)'\cdot cos~x+cos^2x \cdot(-sin~x)=}\\ \\ \mathtt{=((cos~x)'\cdot cos~x+cos~x \cdot (cos~x)')\cdot cos~x-cos^2x\cdot sin~x=}[/tex]

[tex]\displaystyle \mathtt{=(-sin~x \cdot cos~x+cos~x\cdot(-sin~x))\cdot cos~x-cos^2x \cdot sin~x=} \\ \\ \mathtt{=(-2sin~x \cdot cos~x)\cdot cos~x-cos^2x \cdot sin~x=}\\ \\ \mathtt{=-2cos^2x\cdot sin~x-cos^2x \cdot sin~x=-3cos^2x \cdot sin~x}[/tex]

[tex]\displaystyle \mathtt{f''(x)=(f'(x))'=\left(-3cos^2x \cdot sin~x\right)'=-3\left(cos^2x \cdot sin~x\right)'=}\\ \\ \mathtt{=-3\left(\left(cos^2x\right)'\cdot sin~x+cos^2x \cdot (sin~x)'\right)=}\\ \\ \mathtt{-3\left((-2sin~x\cdot cos~x)\cdot sin~x+cos^2x \cdot cos~x\right)=}\\ \\ \mathtt{=-3\left(-2sin^2x\cdot cos~x+cos^3x\right)=6sin^2x \cdot cos~x-3cos^3x}[/tex]

[tex]\displaystyle \mathtt{j)~f(x)=xsin~x+cos~x}\\ \\ \mathtt{f'(x)=(xsin~x+cos~x)'=(xsin~x)'+(cos~x)'=}\\ \\ \mathtt{x'\cdot sin~x+x \cdot (sin~x)'+(-sin~x)=1 \cdot sin~x+x \cdot cos~x-sin~x=}\\ \\ \mathtt{=sin~x+xcos~x-sin~x=xcos~x}\\ \\ \mathtt{f''(x)=(f'(x))'=(xcos~x)'=x' \cdot cos~x+x \cdot (cos~x)'=}\\ \\ \mathtt{=1 \cdot cos~x+x \cdot (-sin~x)=cos~x-xsin~x}[/tex]

[tex]\displaystyle \mathtt{k)~f(x)=x^2 \sqrt{x} ,~x\ \textgreater \ 0}\\ \\ \mathtt{f'(x)=\left(x^2} \sqrt{x} \right)'=\left(x^2\right)'\cdot \sqrt{x} +x^2 \cdot \left( \sqrt{x} \right)'=2x \sqrt{x} +x^2\cdot \frac{1}{2 \sqrt{x} }=} \\ \\ \mathtt{=2x \sqrt{x} + \frac{x^2}{2 \sqrt{x} } = \frac{4x^2+x^2}{2 \sqrt{x} } = \frac{5x^2}{2 \sqrt{x} } }[/tex]

[tex]\displaystyle \mathtt{f''(x)=(f'(x))'=\left( \frac{5x^2}{2 \sqrt{x} } \right)'= \frac{\left(5x^2\right)' \cdot 2 \sqrt{x} -5x^2\cdot \left(2 \sqrt{x} \right)'}{\left(2} \sqrt{x} \right)^2 }=}\\ \\ \mathtt{= \frac{5\left(x^2\right)'\cdot 2 \sqrt{x} -5x^2\cdot 2\left( \sqrt{x} \right)'}{4x}= \frac{5 \cdot 2x \cdot 2 \sqrt{x} -5x^2 \cdot 2 \cdot \frac{1}{2 \sqrt{x} } }{4x}=}[/tex]

[tex]\displaystyle \mathtt{=\frac{20x \sqrt{x}- \frac{5x^2}{ \sqrt{x} }}{4x}= \frac{ \frac{20x^2-5x^2}{\sqrt{x}}}{4x}= \frac{\frac{15x^2}{\sqrt{x}}}{4x}= \frac{15x^2 \sqrt{x} }{x} \cdot \frac{1}{4x}= \frac{15x^2 \sqrt{x} }{4x^2}= \frac{15 \sqrt{x} }{4} }[/tex]
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