Question

Calculaţi derivatele următoarelor funcţii elementare:

Answer 1

$\displaystyle \mathtt{2)~f(x)= \frac{x^2-3x+1}{x^2+4x}}\\ \\ \mathtt{f'(x)=\left( \frac{x^2-3x+1}{x^2+4x} \right)'=}\\ \\ \mathtt{= \frac{(x^2-3x+1)'(x^2+4x)-(x^2-3x+1)(x^2+4x)'}{(x^2+4x)^2}= }\\ \\ \mathtt{= \frac{(2x-3)(x^2+4x)-(x^2-3x+1)(2x+4)}{(x^2+4x)^2}}$

$\displaystyle \mathtt{3)~f(x)=x \cdot ln~x-ln~5\cdot log_5x}\\ \\ \mathtt{f'(x)=(x \cdot ln~x-ln~5\cdot log_5x)'=(x\cdot ln~x)'-(ln~5\cdot log_5x)'=}\\ \\ \mathtt{=x'\cdot ln~x+x \cdot(ln~x)'-(ln~5)'\cdot log_5x+ln~5\cdot(log_5x)'=}\\ \\ \mathtt{=\left(1\cdot ln~x+x \cdot \frac{1}{x}\right)-\left(0\cdot log_5x+ln~5\cdot \frac{1}{xln~x\right)}}=\mathtt{ln~x+1- \frac{1}{x} }$

$\displaystyle \mathtt{4)~f(x)=5^x+3^x\cdot2^x-3^2\cdot2^2}\\ \\ \mathtt{f'(x)=(5^x+3^x\cdot2^x-3^2\cdot2^2)'=(5^x)'+(3^x\cdot2^x)'-(3^2\cdot2^2)'=}\\ \\ \mathtt{=5^xln~5+(3^x)'\cdot2^x+3^x \cdot(2^x)'-0=5^xln~5+3^xln~3\cdot2^x+3^x\cdot2^xln~2}$

$\displaystyle \mathtt{5)~f(x)= \frac{x^3-5x+1}{x^2}}\\ \\ \mathtt{f'(x)=\left( \frac{x^3-5x+1}{x^2}\right)'= \frac{(x^3-5x+1)'\cdot x^2-(x^3-5x+1)\cdot(x^2)'}{(x^2)^2}=}\\ \\ \mathtt{= \frac{(3x^2-5)\cdot x^2-(x^3-5x+1)\cdot(2x)}{x^4}= \frac{3x^4-5x^2-2x^4+10x^2-2x}{x^4}=}\\ \\ \mathtt{= \frac{x^4+5x^2-2x}{x^4}= \frac{x(x^3+5x-2)}{x^4}= \frac{x^3+5x-2}{x^3} }$

$\displaystyle \mathtt{6)~f(x)=(2cos~x-1)(4-3cos~x)}\\ \\ \mathtt{f'(x)=((2cos~x-1)(4-3cos~x))'=}\\ \\ \mathtt{=(2cos~x-1)'(4-3cos~x)+(2cos~x-1)(4-3cos~x)'=}\\ \\ \mathtt{=(-2sin~x)(4-3cos~x)+(2cos~x-1)(3sin~x)=}\\ \\ \mathtt{=-8sin~x+6sin~x\cdot cos~x+6sin~x\cdot cos~x-3sin~x=}\\ \\ \mathtt{=-11sin~x+12sin~x\cdot cos~x}$

$\displaystyle \mathtt{7)~f:R\rightarrow R,~f(x)=x^3+3^x}\\ \\ \mathtt{f'(x)=(x^3+3^x)'=(x^3)'+(3^x)'=3x^2+3^xln~3}$

$\displaystyle \mathtt{8)~f:R\rightarrow R,~f(x)=x^4+7x^2+21x-7}\\ \\ \mathtt{f'(x)=(x^4+7x^2+21x-7)'=(x^4)'+7(x^2)'+21x'-7'=}\\ \\ \mathtt{=4x^3+7\cdot2x+21\cdot1-0=4x^3+14x+21}$

$\displaystyle \mathtt{9)~f:R\rightarrow R,~f(x)=5sin~x-7cos~x}\\ \\ \mathtt{f'(x)=(5sin~x-7cos~x)'=5(sin~x)'-7(cos~x)'=5cos~x+7sin~x}$

$\displaystyle \mathtt{10)~f:(0,\infty)\rightarrow R,~f(x)=x+ln~x}\\ \\ \mathtt{f'(x)=(x+ln~x)'=x'+(ln~x)'=1+ \frac{1}{x} }$

$\displaystyle \mathtt{11)~f:[0,\infty)\rightarrow R,~f(x)=x \cdot \sqrt{x} +x \cdot \sqrt[\mathtt 3]{\mathtt x}}\\ \\ \mathtt{f'(x)=(x \cdot \sqrt{x} +x \cdot \sqrt[\mathtt 3]{\mathtt x})'=(x\sqrt{x} )'+(x\sqrt[\mathtt 3]{\mathtt x})'=}\\ \\ \mathtt{x'\cdot\sqrt{x}+x\cdot(\sqrt{x} )'+x'\cdot\sqrt[\mathtt 3]{\mathtt x}+x\cdot(\sqrt[\mathtt3]{\mathtt x})'=}\\\\\mathtt{=1\cdot\sqrt{x}+x\cdot \frac{1}{2 \sqrt{x}}+1\cdot \sqrt[\mathtt 3]{\mathtt x}+x\cdot\frac{1}{3\sqrt[\mathtt 3]{\mathtt x^2}}=}$
$\displaystyle \mathtt{= \frac{3 \sqrt{x} }{2}+ \frac{4\sqrt[\mathtt 3]{\mathtt x}}{3} }$

$\displaystyle \mathtt{12)~f:(0,\infty)\rightarrow R,~f(x)=x \cdot ln~x}\\ \\ \mathtt{f'(x)=(x \cdot ln~x)'=x' \cdot ln~x+x \cdot (ln~x)'=1 \cdot ln~x+x \cdot \frac{1}{x}=ln~x+1}$

$\displaystyle \mathtt{13)~f:R \rightarrow R,~f(x)=x^3 \cdot 3^x}\\ \\ \mathtt{f'(x)=(x^3 \cdot 3^x)'=(x^3)' \cdot 3^x+x^3 \cdot (3^x)'=3x^2 \cdot 3^x+x^3\cdot3^xln~3}$

$\displaystyle \mathtt{14)~f:R \rightarrow R,~f(x)=x^2 \cdot e^x}\\ \\ \mathtt{f'(x)=(x^2 \cdot e^x)'=(x^2)' \cdot e^x+x^2 \cdot (e^x)'=2x e^x+x^2 e^x}$