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derivative_br_91

91. ddxx3,definition\frac{d}{dx}x^3, definition

ddxx3=lim⁡h→0(x+h)3−x3h=3hx2+3h2x+h3h=lim⁡h→03x2+3hx+h2=3x2 \begin{aligned} &\frac{d}{dx}x^3\\ &=\lim_{h\to0}\frac{(x+h)^3-x^3}{h}=\frac{3hx^2+3h^2x+h^3}{h}\\ &=\lim_{h\to0}3x^2+3hx+h^2=3x^2 \end{aligned}

92. ddx3x+1,def.\frac{d}{dx} \sqrt{3x+1}, def.

ddx3x+1=lim⁡h→03(x+h)+1−3x+1h=lim⁡h→0(3x+3h+1)−(3x+1)h(3x+3h+1+3x+1)=lim⁡h→03(3x+3h+1+3x+1)=323x+1 \begin{aligned} &\frac{d}{dx} \sqrt{3x+1}\\ &=\lim_{h\to0} \frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}\\ &=\lim_{h\to0} \frac{(3x+3h+1)-(3x+1)}{h(\sqrt{3x+3h+1}+\sqrt{3x+1})}\\ &=\lim_{h\to0}\frac{3}{(\sqrt{3x+3h+1}+\sqrt{3x+1})}\\ &=\frac{3}{2\sqrt{3x+1}} \end{aligned}

93. ddx12x+5,def.\frac{d}{dx} \frac{1}{2x+5}, def.

ddx12x+5=lim⁡h→012(x+h)+5−12x+5h=−2h(2x+2h+5)(2x+5)h=lim⁡h→0−2(2x+2h+5)(2x+5)=−2(2x+5)2 \begin{aligned} &\frac{d}{dx} \frac{1}{2x+5}\\ &=\lim_{h\to0} \frac{\frac{1}{2(x+h)+5}-\frac{1}{2x+5}}{h}=\frac{ \frac{-2h}{(2x+2h+5)(2x+5)}}{h}\\ &=\lim_{h\to0} -\frac{2}{(2x+2h+5)(2x+5)}\\ &=-\frac{2}{(2x+5)^2} \end{aligned}

94. ddx1x2,def.\frac{d}{dx}\frac{1}{x^2}, def.

ddx1x2=lim⁡h→01(x+h)2−1x2h=−2xh−h2x2(x+h)2h=−2x−hx2(x+h)2=−2xx4=−2x3 \begin{aligned} &\frac{d}{dx}\frac{1}{x^2}\\ &=\lim_{h\to0} \frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}= \frac{\frac{-2xh-h^2}{x^2(x+h)^2}}{h}=\frac{-2x-h}{x^2(x+h)^2}\\ &=\frac{-2x}{x^4}=-\frac{2}{x^3} \end{aligned}

95. ddxsinx,def.\frac{d}{dx}sinx, def.

ddxsinx=lim⁡h→0sin(x+h)−sin(x)h=lim⁡h→0sin(x)cos(h)+cos(x)sin(h)−sin(x)h=lim⁡h→0sinx(cos(h)−1)h+cos(x)sin(h)hcos(h)=1−h22!+h44!−...lim⁡h→0cos(h)−1h=hc+h3c+..=O(h)=0sin(h)=h−h33!+...lim⁡h→0sin(h)h=1−O(h2)=1∴=sin(x)0+cos(x)1=cos(x) \begin{aligned} &\frac{d}{dx} sinx=\lim_{h\to0} \frac{sin(x+h)-sin(x)}{h}\\ &=\lim_{h\to0} \frac{ sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\\ &=\lim_{h\to0} \frac{sinx(cos(h)-1)}{h}+cos(x)\frac{sin(h)}{h}\\ & cos(h)=1-\frac{h^2}{2!}+\frac{h^4}{4!}-...\\ & \lim_{h\to0} \frac{cos(h)-1}{h}=hc+h^3c+..=O(h)=0 \\ & sin(h)=h-\frac{h^3}{3!}+...\\ & \lim_{h\to0} \frac{sin(h)}{h}=1-O(h^2)=1\\ & \therefore =sin(x)0+cos(x)1=cos(x) \end{aligned}

96. ddxsecx,def.\frac{d}{dx}secx, def.

ddxsec(x)=lim⁡h→0sec⁡(x+h)−sec⁡(x)h=lim⁡h→01/cos⁡(x+h)−1/cos⁡(x)h=cos(x)−cos(x+h)cos(x+h)cos(x)h=cos(x)−cos(x)cos(h)+sin(x)sin(h)(cos(x)cos(h)−sin(x)sin(h))cos(x)h=1−cos(h)+tan(x)sin(h)h(cos(x)cos(h)−sin(x)sin(h))=lim⁡1cos(x)cos(h)−sin(x)sin(h)(lim⁡1−cos(h)h+lim⁡tan(x)sin(h)h)=1cos(x)(0+tan(x))=sec(x)tan(x) \begin{aligned} &\frac{d}{dx}sec(x)=\lim_{h\to0}\frac{\sec(x+h)-\sec(x)}{h}\\ &=\lim_{h\to0}\frac{1/\cos(x+h)-1/\cos(x)}{h}=\frac{cos(x)-cos(x+h)}{cos(x+h)cos(x)h}\\ &=\frac{cos(x)-cos(x)cos(h)+sin(x)sin(h)}{(cos(x)cos(h)-sin(x)sin(h))cos(x)h}\\ &=\frac{1-cos(h)+tan(x)sin(h)}{h(cos(x)cos(h)-sin(x)sin(h))}\\ &=\lim \frac{1}{cos(x)cos(h)-sin(x)sin(h)} (\lim \frac{1-cos(h)}{h}+\lim \frac{tan(x)sin(h)}{h}) \\ &=\frac{1}{cos(x)}(0+tan(x))=sec(x)tan(x) \end{aligned}

97. ddxarcsinx,def.\frac{d}{dx}arcsinx, def.

ddxarcsinx=lim⁡h→0arcsin(x+h)−arcsin(x)h \begin{aligned} &\frac{d}{dx}arcsinx=\lim_{h\to0} \frac{arcsin(x+h)-arcsin(x)}{h}\\ \end{aligned}

fail

ddxarcsinx=lim⁡h→0arcsin(x+h)−arcsin(x)hsin(a−b)=sinacosb−sinbcosaarcsin⁡sin(a−b)=arcsin⁡(sinacosb−sinbcosa)a−b=arcsin⁡(sinacosb−sinbcosa)a=arcsin(x+h),b=arcsin(x)arcsin(x+h)−arcsin(x)=arcsin(sin(arcsin(x+h))cos(arcsin(x))−sin(arcsin(x))cos(arcsin(x+h)))=arcsin((x+h)cos(arcsin(x))−xcos(arcsin(x+h)))(cos(x)=1−sin2(x))We know,lim⁡x→0sinxx=1,lim⁡x→0sin(x)=lim⁡x→0xSo,lim⁡x→0sin−1(x)=lim⁡x→0x=arcsin((x+h)1−x2−x1−(x+h)2)ddxarcsinx=lim⁡h→0arcsin(x+h)−arcsin(x)h=lim⁡h→0arcsin((x+h)1−x2−x1−(x+h)2)h=lim⁡h→0(x+h)1−x2−x1−(x+h)2h=lim⁡h→0(x+h)2(1−x2)−x2(1−(x+h)2)h((x+h)1−x2+x1−(x+h)2)Numerator=(x+h)2−x2(x+h)2−x2+x2(x+h)2Numerator=(x+h)2−x2=2xh+h2=lim⁡h→02x+h(x+h)1−x2+x1−(x+h)2=2xx1−x2+x1−x2=2x2x1−x2=11−x2 \begin{aligned} &\frac{d}{dx}arcsinx=\lim_{h\to0} \frac{arcsin(x+h)-arcsin(x)}{h}\\ & sin(a-b) = sinacosb-sinbcosa\\ & \arcsin {sin(a-b)} = \arcsin {(sinacosb-sinbcosa)}\\ & a-b = \arcsin {(sinacosb-sinbcosa)}\\ & a=arcsin(x+h), b=arcsin(x)\\ &arcsin(x+h)-arcsin(x) = arcsin(sin(arcsin(x+h))cos(arcsin(x))-sin(arcsin(x))cos(arcsin(x+h)))\\ &=arcsin((x+h)cos(arcsin(x))-xcos(arcsin(x+h)))\\ & (cos(x) = \sqrt {1-sin^2(x)}) \\ & \text{We know,} \lim_{x\to0} \frac{sin x}{x}=1, \lim _{x\to0} sin(x) = \lim_{x\to0} x\\ &So, \lim_{x\to0} sin^{-1}(x)=\lim_{x\to0}x &=arcsin( (x+h) \sqrt{1-x^2}-x\sqrt{1-(x+h)^2} )\\ &\frac{d}{dx}arcsinx=\lim_{h\to0} \frac{arcsin(x+h)-arcsin(x)}{h}\\ &=\lim_{h\to0} \frac{arcsin( (x+h) \sqrt{1-x^2}-x\sqrt{1-(x+h)^2} )}{h}\\ &=\lim_{h\to0} \frac{ (x+h) \sqrt{1-x^2}-x\sqrt{1-(x+h)^2} }{h} \\ &=\lim_{h\to0} \frac{ (x+h)^2(1-x^2)-x^2(1-(x+h)^2)} {h ((x+h) \sqrt{1-x^2}+x\sqrt{1-(x+h)^2} )} \\ &Numerator=(x+h)^2-x^2(x+h)^2-x^2+x^2(x+h)^2\\ &Numerator=(x+h)^2-x^2=2xh+h^2\\ &=\lim_{h\to0} \frac{ 2x+h} {(x+h) \sqrt{1-x^2}+x\sqrt{1-(x+h)^2} } \\ &=\frac{2x}{x\sqrt{1-x^2}+x\sqrt{1-x^2}}=\frac{2x}{2x\sqrt{1-x^2}}\\ &=\frac{1}{\sqrt{1-x^2}} \end{aligned}

98. ddxarctanx,def.\frac{d}{dx}arctanx, def.

Try like upper case, lim tan(x)/x = 1, atan(x)/x=1

ddxarctan(x)=lim⁡h→0arctan(x+h)−arctan(x)htan(a−b)=tan(a)−tan(b)1+tan(a)tan(b)a−b=arctan(tan(a)−tan(b)1+tan(a)tan(b))Numerator=arctan(x+h)−arctan(x)N=arctan(tan(arctan(x+h))−tan(arctan(x))1+tan(arctan(x+h))tan(arctan(x)))=arctan(x+h−x1+(x+h)x)=artan(h1+x2+hx)So,ddxarctan(x)=lim⁡h→0arctan(x+h)−arctan(x)h=lim⁡h→0artan(h1+x2+hx)h=lim⁡h→011+x2+hx=11+x2 \begin{aligned} &\frac{d}{dx}arctan(x)=\lim_{h\to0} \frac{arctan(x+h)-arctan(x)}{h}\\ & tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)tan(b)}\\ & a-b = arctan(\frac{tan(a)-tan(b)}{1+tan(a)tan(b)})\\ &Numerator=arctan(x+h)-arctan(x)\\ &N=arctan( \frac{tan(arctan(x+h))-tan(arctan(x))}{1+tan(arctan(x+h))tan(arctan(x))} )\\ &=arctan( \frac{x+h-x}{1+(x+h)x} )=artan(\frac{h}{1+x^2+hx})\\ &So, \\ &\frac{d}{dx}arctan(x)=\lim_{h\to0} \frac{arctan(x+h)-arctan(x)}{h}\\ &=\lim_{h\to0} \frac{artan(\frac{h}{1+x^2+hx})}{h}\\ &=\lim_{h\to0} \frac{1}{1+x^2+hx}=\frac{1}{1+x^2}\\ \end{aligned}

99. ddxf(x)g(x),def.\frac{d}{dx}f(x)g(x), def.

ddxf(x)g(x)=lim⁡h→0f(x+h)g(x+h)−f(x)g(x)h=lim⁡h→0f(x+h)g(x+h)−f(x)g(x)−g(x+h)f(x)+g(x+h)f(x)h=lim⁡h→0g(x+h)(f(x+h)−f(x))+f(x)(g(x+h)−g(x))h=lim⁡h→0g(x+h)f(x+h)−f(x)h+f(x)g(x+h)−g(x)h=g(x)f′(x)+f(x)g′(x)=f′(x)g(x)+f(x)g′(x) \begin{aligned} &\frac{d}{dx}f(x)g(x)=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)-g(x+h)f(x)+g(x+h)f(x)}{h}\\ &=\lim_{h\to0} \frac{g(x+h)(f(x+h)-f(x))+f(x)(g(x+h)-g(x))}{h}\\ &=\lim_{h\to0} g(x+h)\frac{f(x+h)-f(x)}{h}+f(x)\frac{g(x+h)-g(x)}{h}\\ &=g(x)f'(x)+f(x)g'(x)=f'(x)g(x)+f(x)g'(x) \end{aligned}

100. ddxf(x)g(x),def.\frac{d}{dx} \frac{f(x)}{g(x)}, def.

ddxf(x)g(x)=lim⁡h→0f(x+h)g(x+h)−f(x)g(x)h=f(x+h)g(x)−f(x)g(x+h)g(x)g(x+h)h=f(x+h)g(x)−f(x)g(x+h)−g(x)f(x)+g(x)f(x)hg(x)g(x+h)=g(x)(f(x+h)−f(x))−f(x)(g(x+h)−g(x))hg(x)g(x+h)=g(x)f′(x)−f(x)g′(x)g(x)2 \begin{aligned} &\frac{d}{dx} \frac{f(x)}{g(x)}=\lim_{h\to0}\frac{ \frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h} =\frac{ \frac{f(x+h)g(x)-f(x)g(x+h)}{g(x)g(x+h)} }{h}\\ &=\frac{ f(x+h)g(x)-f(x)g(x+h)-g(x)f(x)+g(x)f(x)}{hg(x)g(x+h)}\\ &=\frac{g(x)(f(x+h)-f(x))-f(x)(g(x+h)-g(x))}{hg(x)g(x+h)}\\ &=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2} \end{aligned}

101. ddxxxx\frac{d}{dx} x^{{x}^{x}}

First,ddxxxy=xx,lny=xlnx,(1/y)y′=lnx+x(1/x)y′=ylnx+y=xxlnx+xx \frac{d}{dx}x^x\\ y=x^x, lny=xlnx, (1/y)y'=lnx+x(1/x)\\ y'=ylnx+y=x^xlnx+x^x
Second,
ddxxxxy=xxxlny=xxln(x)1yy′=(xxlnx+xx)ln(x)+xx1x=xx((lnx)2+ln(x)+1x)y′=xxxxx((lnx)2+ln(x)+1x) \begin{aligned} &\frac{d}{dx} x^{x^{x}}\\ &y=x^{x^{x}} \\ &ln y = x^x ln(x) \\ &\frac{1}{y} y' = (x^xlnx+x^x)ln(x)+x^x\frac{1}{x}\\ &=x^x((lnx)^2+ln(x)+\frac{1}{x})\\ &y'=x^{x^x}x^x((lnx)^2+ln(x)+\frac{1}{x})\\ \end{aligned}

The END

Author: crazyj7@gmail.com

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