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Question Number 11920 by carrot last updated on 05/Apr/17

$$differentiateeachfunctionfromfirstprinciple$$$$1)f\left(x\right)=1+\frac{1}{x}$$$$2)f\left(x\right)=\frac{1}{2x+3}$$$$3)f\left(x\right)=sin2x$$$$4)f\left(x\right)=co2x$$$$$$

Answered by sandy_suhendra last updated on 05/Apr/17

$$1)\mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^{-1}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=-{\mathrm{x}}^{-2}=\frac{-1}{{\mathrm{x}}^2}$$$$2)\mathrm{f}\left(\mathrm{x}\right)={(2\mathrm{x}+3)}^{-1}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=-1{(2\mathrm{x}+3)}^{-2}\left(2\right)=\frac{-2}{{(2\mathrm{x}+3)}^2}$$$$3)\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=2\cos 2\mathrm{x}$$$$4)\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=-2\sin 2\mathrm{x}$$

Commented by carrot last updated on 05/Apr/17

$$ohwhatyoudidwasdifferentiateitandthatsnotwhattheyaskedfor$$

Commented by sandy_suhendra last updated on 06/Apr/17

$$\mathrm{oh}\mathrm{sorry}\mathrm{because}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{with}\mathrm{the}\mathrm{question}$$$$\mathrm{but}\mathrm{I}\mathrm{have}\mathrm{fixed}\mathrm{for}\mathrm{no}3\mathrm{and}4$$

Answered by FilupS last updated on 05/Apr/17

$$\left(1\right)$$$$f{}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$$$$f{}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{(1+\frac{1}{x+h})-(1+\frac{1}{x})}{h}$$$$=\underset{h\to 0}{\lim }\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$$$$=\underset{h\to 0}{\lim }\frac{\left(\frac{x-x-h}{x^2+xh}\right)}{h}$$$$=\underset{h\to 0}{\lim }\frac{\left(\frac{-h}{x^2+xh}\right)}{h}$$$$=\underset{h\to 0}{\lim }(-\frac{1}{x^2+xh})$$$$=-\frac{1}{x^2}$$

Answered by FilupS last updated on 05/Apr/17

$$\left(2\right)$$$$f{}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$$$$f{}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{\left(\frac{1}{2x+2h+3}\right)-\left(\frac{1}{2x+3}\right)}{h}$$$$=\underset{h\to 0}{\lim }\frac{(\frac{1}{2x+2h+3}-\frac{1}{2x+3})}{h}$$$$=\underset{h\to 0}{\lim }\frac{\left(\frac{2x+3-2x-2h-3}{(2x+2h+3)(2x+3)}\right)}{h}$$$$=\underset{h\to 0}{\lim }\frac{\left(\frac{-2h}{(4x^2+6x+4hx+6h+6x+9)}\right)}{h}$$$$=\underset{h\to 0}{\lim }\frac{\left(\frac{-2}{(4x^2+6x+4hx+6h+6x+9)}\right)}{1}$$$$=-\frac{2}{4x^2+12x+9}$$$$=-\frac{2}{{(2x+3)}^2}$$

Answered by sandy_suhendra last updated on 06/Apr/17

$$3)\mathrm{li}\underset{\mathrm{h}\to 0}{\mathrm{m}}\frac{\sin 2(\mathrm{x}+\mathrm{h})-\sin 2\mathrm{x}}{\mathrm{h}}$$$$=\mathrm{li}\underset{\mathrm{h}\to 0}{\mathrm{m}}\frac{2\cos (2\mathrm{x}+\mathrm{h})\sin \mathrm{h}}{\mathrm{h}}$$$$=\mathrm{li}\underset{\mathrm{h}\to 0}{\mathrm{m}}2\cos (2\mathrm{x}+\mathrm{h})\frac{\sin \mathrm{h}}{\mathrm{h}}$$$$=2\cos (2\mathrm{x}+0).1=2\cos 2\mathrm{x}$$$$$$$$4)\mathrm{li}\underset{\mathrm{h}\to 0}{\mathrm{m}}\frac{\cos 2(\mathrm{x}+\mathrm{h})-\cos 2\mathrm{x}}{\mathrm{h}}$$$$=\mathrm{li}\underset{\mathrm{x}\to 0}{\mathrm{m}}\frac{-2\sin (2\mathrm{x}+\mathrm{h})\sin \mathrm{h}}{\mathrm{h}}$$$$=\mathrm{li}\underset{\mathrm{x}\to 0}{\mathrm{m}}-2\sin (2\mathrm{x}+\mathrm{h})\frac{\sin \mathrm{h}}{\mathrm{h}}$$$$=-2\sin (2\mathrm{x}+0).1=-2\sin 2\mathrm{x}$$

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