Mathematics Guide

M2 Guide ​

First Principles ​

Powers ​

Find ddxx3\dfrac{d}{dx}x^{3} from first principles.

Solution

ddxx3=lim⁡h→0(x+h)3−x3h=lim⁡h→0(x3+3x2h+3xh2+h3)−x3h=lim⁡h→03x2h+3xh2+h3h=lim⁡h→0(3x2+3xh+h2)=3x2+3x(0)+(0)2=3x2\begin{aligned} \dfrac{d}{dx}x^{3} &= \lim_{h \to 0}\dfrac{(x+h)^{3}-x^{3}}{h} \\ &= \lim_{h \to 0}\dfrac{(x^{3}+3x^{2}h+3xh^{2}+h^{3})-x^{3}}{h} \\ &= \lim_{h \to 0}\dfrac{3x^{2}h+3xh^{2}+h^{3}}{h} \\ &= \lim_{h \to 0}(3x^{2}+3xh+h^{2}) \\ &= 3x^{2}+3x(0)+(0)^{2} \\ &= 3x^{2} \end{aligned}

Find ddx1x+1\dfrac{d}{dx}\dfrac{1}{x+1} from first principles.

Solution

ddx1x+1=lim⁡h→01h[1x+h+1−1x+1]=lim⁡h→01h[(x+1)−(x+h+1)(x+h+1)(x+1)]=lim⁡h→01h[−h(x+h+1)(x+1)]=lim⁡h→0−1(x+h+1)(x+1)=−1(x+0+1)(x+1)=−1(x+1)2\begin{aligned} \dfrac{d}{dx}\dfrac{1}{x+1} &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{1}{x+h+1}-\dfrac{1}{x+1}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{-h}{(x+h+1)(x+1)}\right] \\ &= \lim_{h \to 0}\dfrac{-1}{(x+h+1)(x+1)} \\ &= \dfrac{-1}{(x+0+1)(x+1)} \\ &= \dfrac{-1}{(x+1)^{2}} \end{aligned}

Find ddxx\dfrac{d}{dx}\sqrt{x} from first principles.

Solution

ddxx=lim⁡h→0x+h−xh=lim⁡h→0(x+h−x)(x+h+x)h(x+h+x)=lim⁡h→0(x+h)2−(x)2h(x+h+x)=lim⁡h→0x+h−xh(x+h+x)=lim⁡h→0hh(x+h+x)=lim⁡h→01x+h+x=1x+0+x=12x\begin{aligned} \dfrac{d}{dx}\sqrt{x} &= \lim_{h \to 0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h} \\ &= \lim_{h \to 0}\dfrac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \to 0}\dfrac{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \to 0}\dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \to 0}\dfrac{h}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \to 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}} \\ &= \dfrac{1}{\sqrt{x+0}+\sqrt{x}} \\ &= \dfrac{1}{2\sqrt{x}} \end{aligned}

Find ddx1x\dfrac{d}{dx}\dfrac{1}{\sqrt{x}} from first principles.

Solution

ddx1x=lim⁡h→01h[1x+h−1x]=lim⁡h→01h[x−x+hx+hx]=lim⁡h→01h[(x−x+h)(x+x+h)x+hx(x+x+h)]=lim⁡h→01h[(x)2−(x+h)2x+hx(x+x+h)]=lim⁡h→01h[x−(x+h)x+hx(x+x+h)]=lim⁡h→01h[−hx+hx(x+x+h)]=lim⁡h→0−1x+hx(x+x+h)=−1x+0x(x+x+0)=−12xx\begin{aligned} \dfrac{d}{dx}\dfrac{1}{\sqrt{x}} &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{1}{\sqrt{x+h}}-\dfrac{1}{\sqrt{x}}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x+h}\sqrt{x}}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{(\sqrt{x}-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{(\sqrt{x})^{2}-(\sqrt{x+h})^{2}}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{x-(x+h)}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{-h}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})}\right] \\ &= \lim_{h \to 0}\dfrac{-1}{\sqrt{x+h}\sqrt{x}(\sqrt{x}+\sqrt{x+h})} \\ &= \dfrac{-1}{\sqrt{x+0}\sqrt{x}(\sqrt{x}+\sqrt{x+0})} \\ &= \dfrac{-1}{2x\sqrt{x}} \end{aligned}

Find ddxx3\dfrac{d}{dx}\sqrt[3]{x} from first principles.

Solution

Let a=x+h3a=\sqrt[3]{x+h} and b=x3b=\sqrt[3]{x}.

ddxx3=lim⁡h→0x+h3−x3h=lim⁡h→0a−bh=lim⁡h→0(a−b)(a2+ab+b2)h(a2+ab+b2)=lim⁡h→0a3−b3h(a2+ab+b2)=lim⁡h→0(x+h3)3−(x3)3h[(x+h3)2+x+h3x3+(x3)2]=lim⁡h→0x+h−xh[(x+h3)2+x+h3x3+(x3)2]=lim⁡h→01(x+h3)2+x+h3x3+(x3)2=1(x3)2+x3x3+(x3)2=13(x3)2\begin{aligned} \dfrac{d}{dx}\sqrt[3]{x} &= \lim_{h \to 0}\dfrac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} \\ &= \lim_{h \to 0}\dfrac{a-b}{h} \\ &= \lim_{h \to 0}\dfrac{(a-b)(a^{2}+ab+b^{2})}{h(a^{2}+ab+b^{2})} \\ &= \lim_{h \to 0}\dfrac{a^{3}-b^{3}}{h(a^{2}+ab+b^{2})} \\ &= \lim_{h \to 0}\dfrac{(\sqrt[3]{x+h})^{3}-(\sqrt[3]{x})^{3}}{h\left[(\sqrt[3]{x+h})^{2}+\sqrt[3]{x+h}\sqrt[3]{x}+(\sqrt[3]{x})^{2}\right]} \\ &= \lim_{h \to 0}\dfrac{x+h-x}{h\left[(\sqrt[3]{x+h})^{2}+\sqrt[3]{x+h}\sqrt[3]{x}+(\sqrt[3]{x})^{2}\right]} \\ &= \lim_{h \to 0}\dfrac{1}{(\sqrt[3]{x+h})^{2}+\sqrt[3]{x+h}\sqrt[3]{x}+(\sqrt[3]{x})^{2}} \\ &= \dfrac{1}{(\sqrt[3]{x})^{2}+\sqrt[3]{x}\sqrt[3]{x}+(\sqrt[3]{x})^{2}} \\ &= \dfrac{1}{3(\sqrt[3]{x})^{2}} \end{aligned}

Trig ​

Find ddxsin⁡x\dfrac{d}{dx}\sin x from first principles.

Method 1: Sum-to-product Formula

ddxsin⁡x=lim⁡h→0sin⁡(x+h)−sin⁡xh=lim⁡h→02cos⁡(x+h+x2)sin⁡(x+h−x2)h←sum-to-product=lim⁡h→0cos⁡(x+h2)sin⁡(h2)h2=lim⁡h→0cos⁡(x+h2)⋅sin⁡(h2)h2=cos⁡(x+0)⋅(1)=cos⁡x\begin{aligned} \dfrac{d}{dx}\sin x &= \lim_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h} \\ &= \lim_{h \to 0}\dfrac{2\cos\left(\dfrac{x+h+x}{2}\right)\sin\left(\dfrac{x+h-x}{2}\right)}{h} && \leftarrow \text{sum-to-product} \\ &= \lim_{h \to 0}\dfrac{\cos\left(x+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} \\ &= \lim_{h \to 0}\cos\left(x+\dfrac{h}{2}\right)\cdot\dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} \\ &= \cos(x+0)\cdot(1) \\ &= \cos x \end{aligned}

Method 2: Compound Angle Formula

ddxsin⁡x=lim⁡h→0sin⁡(x+h)−sin⁡xh=lim⁡h→0sin⁡xcos⁡h+cos⁡xsin⁡h−sin⁡xh=lim⁡h→0cos⁡xsin⁡h−sin⁡x(1−cos⁡h)h=lim⁡h→0[cos⁡x⋅sin⁡hh−sin⁡x⋅1−cos⁡hh]=lim⁡h→0[cos⁡x⋅sin⁡hh−sin⁡x⋅2sin⁡2(h2)h]=lim⁡h→0[cos⁡x⋅sin⁡hh−sin⁡xsin⁡(h2)⋅sin⁡(h2)h2]=cos⁡x⋅(1)−sin⁡xsin⁡0⋅(1)=cos⁡x\begin{aligned} \dfrac{d}{dx}\sin x &= \lim_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h} \\ &= \lim_{h \to 0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h} \\ &= \lim_{h \to 0}\dfrac{\cos x\sin h-\sin x(1-\cos h)}{h} \\ &= \lim_{h \to 0}\left[\cos x\cdot\dfrac{\sin h}{h}-\sin x\cdot\dfrac{1-\cos h}{h}\right] \\ &= \lim_{h \to 0}\left[\cos x\cdot\dfrac{\sin h}{h}-\sin x\cdot\dfrac{2\sin^{2}\left(\dfrac{h}{2}\right)}{h}\right] \\ &= \lim_{h \to 0}\left[\cos x\cdot\dfrac{\sin h}{h}-\sin x\sin\left(\dfrac{h}{2}\right)\cdot\dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right] \\ &= \cos x\cdot(1)-\sin x\sin 0\cdot(1) \\ &= \cos x \end{aligned}

Find ddxcos⁡x\dfrac{d}{dx}\cos x from first principles.

Method 1: Sum-to-product Formula

ddxcos⁡x=lim⁡h→0cos⁡(x+h)−cos⁡xh=lim⁡h→0−2sin⁡(x+h+x2)sin⁡(x+h−x2)h←sum-to-product=lim⁡h→0−sin⁡(x+h2)sin⁡(h2)h2=lim⁡h→0−sin⁡(x+h2)⋅sin⁡(h2)h2=−sin⁡(x+0)⋅(1)=−sin⁡x\begin{aligned} \dfrac{d}{dx}\cos x &= \lim_{h \to 0}\dfrac{\cos(x+h)-\cos x}{h} \\ &= \lim_{h \to 0}\dfrac{-2\sin\left(\dfrac{x+h+x}{2}\right)\sin\left(\dfrac{x+h-x}{2}\right)}{h} && \leftarrow \text{sum-to-product} \\ &= \lim_{h \to 0}\dfrac{-\sin\left(x+\dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} \\ &= \lim_{h \to 0}-\sin\left(x+\dfrac{h}{2}\right)\cdot\dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}} \\ &= -\sin(x+0)\cdot(1) \\ &= -\sin x \end{aligned}

Method 2: Compound Angle Formula

ddxcos⁡x=lim⁡h→0cos⁡(x+h)−cos⁡xh=lim⁡h→0cos⁡xcos⁡h−sin⁡xsin⁡h−cos⁡xh=lim⁡h→0−sin⁡xsin⁡h−cos⁡x(1−cos⁡h)h=lim⁡h→0[−sin⁡x⋅sin⁡hh−cos⁡x⋅1−cos⁡hh]=lim⁡h→0[−sin⁡x⋅sin⁡hh−cos⁡x⋅2sin⁡2(h2)h]=lim⁡h→0[−sin⁡x⋅sin⁡hh−cos⁡xsin⁡(h2)⋅sin⁡(h2)h2]=−sin⁡x⋅(1)−cos⁡xsin⁡0⋅(1)=−sin⁡x\begin{aligned} \dfrac{d}{dx}\cos x &= \lim_{h \to 0}\dfrac{\cos(x+h)-\cos x}{h} \\ &= \lim_{h \to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h} \\ &= \lim_{h \to 0}\dfrac{-\sin x\sin h-\cos x(1-\cos h)}{h} \\ &= \lim_{h \to 0}\left[-\sin x\cdot\dfrac{\sin h}{h}-\cos x\cdot\dfrac{1-\cos h}{h}\right] \\ &= \lim_{h \to 0}\left[-\sin x\cdot\dfrac{\sin h}{h}-\cos x\cdot\dfrac{2\sin^{2}\left(\dfrac{h}{2}\right)}{h}\right] \\ &= \lim_{h \to 0}\left[-\sin x\cdot\dfrac{\sin h}{h}-\cos x\sin\left(\dfrac{h}{2}\right)\cdot\dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right] \\ &= -\sin x\cdot(1)-\cos x\sin 0\cdot(1) \\ &= -\sin x \end{aligned}

Find ddxtan⁡x\dfrac{d}{dx}\tan x from first principles.

Method 1

ddxtan⁡x=lim⁡h→01h[tan⁡(x+h)−tan⁡x]=lim⁡h→01h[sin⁡(x+h)cos⁡(x+h)−sin⁡xcos⁡x]=lim⁡h→01h[sin⁡(x+h)cos⁡x−sin⁡xcos⁡(x+h)cos⁡(x+h)cos⁡x]=lim⁡h→01h[sin⁡(x+h−x)cos⁡(x+h)cos⁡x]=lim⁡h→01h[sin⁡hcos⁡(x+h)cos⁡x]=lim⁡h→0[sin⁡hh⋅1cos⁡(x+h)cos⁡x]=(1)⋅1cos⁡(x+0)cos⁡x=1cos⁡2x=sec⁡2x\begin{aligned} \dfrac{d}{dx}\tan x &= \lim_{h \to 0}\dfrac{1}{h}\left[\tan(x+h)-\tan x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\sin(x+h)}{\cos(x+h)}-\dfrac{\sin x}{\cos x}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\sin(x+h)\cos x-\sin x\cos(x+h)}{\cos(x+h)\cos x}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\sin(x+h-x)}{\cos(x+h)\cos x}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\sin h}{\cos(x+h)\cos x}\right] \\ &= \lim_{h \to 0}\left[\dfrac{\sin h}{h}\cdot\dfrac{1}{\cos(x+h)\cos x}\right] \\ &= (1)\cdot\dfrac{1}{\cos(x+0)\cos x} \\ &= \dfrac{1}{\cos^{2}x} \\ &= \sec^{2}x \end{aligned}

Method 2

ddxtan⁡x=lim⁡h→01h[tan⁡(x+h)−tan⁡x]=lim⁡h→01h[tan⁡(x+h)−tan⁡x1+tan⁡(x+h)tan⁡x][1+tan⁡(x+h)tan⁡x]=lim⁡h→01h[tan⁡(x+h−x)][1+tan⁡(x+h)tan⁡x]=lim⁡h→01h[tan⁡h][1+tan⁡(x+h)tan⁡x]=lim⁡h→0sin⁡hh⋅1cos⁡h[1+tan⁡(x+h)tan⁡x]=(1)⋅1cos⁡0[1+tan⁡(x+0)tan⁡x]=1+tan⁡2x=sec⁡2x\begin{aligned} \dfrac{d}{dx}\tan x &= \lim_{h \to 0}\dfrac{1}{h}\left[\tan(x+h)-\tan x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\tan(x+h)-\tan x}{1+\tan(x+h)\tan x}\right]\left[1+\tan(x+h)\tan x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\tan(x+h-x)\right]\left[1+\tan(x+h)\tan x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\tan h\right]\left[1+\tan(x+h)\tan x\right] \\ &= \lim_{h \to 0}\dfrac{\sin h}{h}\cdot\dfrac{1}{\cos h}\left[1+\tan(x+h)\tan x\right] \\ &= (1)\cdot\dfrac{1}{\cos 0}\left[1+\tan(x+0)\tan x\right] \\ &= 1+\tan^{2}x \\ &= \sec^{2}x \end{aligned}

Find ddxcsc⁡x\dfrac{d}{dx}\csc x from first principles.

Solution

ddxcsc⁡x=lim⁡h→01h[csc⁡(x+h)−csc⁡x]=lim⁡h→01h[1sin⁡(x+h)−1sin⁡x]=lim⁡h→01h[sin⁡x−sin⁡(x+h)sin⁡(x+h)sin⁡x]=lim⁡h→0sin⁡(x+h)−sin⁡xh×−1sin⁡(x+h)sin⁡x=⋯(same steps as ddxsin⁡x=lim⁡h→0sin⁡(x+h)−sin⁡xh)=cos⁡x×−1sin⁡(x+0)sin⁡x=−cos⁡xsin⁡2x=−csc⁡xcot⁡x\begin{aligned} \dfrac{d}{dx}\csc x &= \lim_{h \to 0}\dfrac{1}{h}\left[\csc(x+h)-\csc x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{1}{\sin(x+h)}-\dfrac{1}{\sin x}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\sin x-\sin(x+h)}{\sin(x+h)\sin x}\right] \\ &= \lim_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h}\times\dfrac{-1}{\sin(x+h)\sin x} \\ &= \cdots\left(\text{same steps as }\dfrac{d}{dx}\sin x=\lim_{h \to 0}\dfrac{\sin(x+h)-\sin x}{h}\right) \\ &= \cos x\times\dfrac{-1}{\sin(x+0)\sin x} \\ &= \dfrac{-\cos x}{\sin^{2}x} \\ &= -\csc x\cot x \end{aligned}

Find ddxsec⁡x\dfrac{d}{dx}\sec x from first principles.

Solution

ddxsec⁡x=lim⁡h→01h[sec⁡(x+h)−sec⁡x]=lim⁡h→01h[1cos⁡(x+h)−1cos⁡x]=lim⁡h→01h[cos⁡x−cos⁡(x+h)cos⁡(x+h)cos⁡x]=lim⁡h→0cos⁡(x+h)−cos⁡xh×−1cos⁡(x+h)cos⁡x=⋯(same steps as ddxcos⁡x=lim⁡h→0cos⁡(x+h)−cos⁡xh)=−sin⁡x×−1cos⁡(x+0)cos⁡x=sin⁡xcos⁡2x=sec⁡xtan⁡x\begin{aligned} \dfrac{d}{dx}\sec x &= \lim_{h \to 0}\dfrac{1}{h}\left[\sec(x+h)-\sec x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{1}{\cos(x+h)}-\dfrac{1}{\cos x}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\cos x-\cos(x+h)}{\cos(x+h)\cos x}\right] \\ &= \lim_{h \to 0}\dfrac{\cos(x+h)-\cos x}{h}\times\dfrac{-1}{\cos(x+h)\cos x} \\ &= \cdots\left(\text{same steps as }\dfrac{d}{dx}\cos x=\lim_{h \to 0}\dfrac{\cos(x+h)-\cos x}{h}\right) \\ &= -\sin x\times\dfrac{-1}{\cos(x+0)\cos x} \\ &= \dfrac{\sin x}{\cos^{2}x} \\ &= \sec x\tan x \end{aligned}

Find ddxcot⁡x\dfrac{d}{dx}\cot x from first principles.

Solution

ddxcot⁡x=lim⁡h→01h[cot⁡(x+h)−cot⁡x]=lim⁡h→01h[1tan⁡(x+h)−1tan⁡x]=lim⁡h→01h[tan⁡x−tan⁡(x+h)tan⁡(x+h)tan⁡x]=lim⁡h→0tan⁡(x+h)−tan⁡xh×−1tan⁡(x+h)tan⁡x=⋯(same steps as ddxtan⁡x=lim⁡h→0tan⁡(x+h)−tan⁡xh)=sec⁡2x×−1tan⁡(x+0)tan⁡x=1cos⁡2x×−1tan⁡2x=−1sin⁡2x=−csc⁡2x\begin{aligned} \dfrac{d}{dx}\cot x &= \lim_{h \to 0}\dfrac{1}{h}\left[\cot(x+h)-\cot x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{1}{\tan(x+h)}-\dfrac{1}{\tan x}\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\left[\dfrac{\tan x-\tan(x+h)}{\tan(x+h)\tan x}\right] \\ &= \lim_{h \to 0}\dfrac{\tan(x+h)-\tan x}{h}\times\dfrac{-1}{\tan(x+h)\tan x} \\ &= \cdots\left(\text{same steps as }\dfrac{d}{dx}\tan x=\lim_{h \to 0}\dfrac{\tan(x+h)-\tan x}{h}\right) \\ &= \sec^{2}x\times\dfrac{-1}{\tan(x+0)\tan x} \\ &= \dfrac{1}{\cos^{2}x}\times\dfrac{-1}{\tan^{2}x} \\ &= \dfrac{-1}{\sin^{2}x} \\ &= -\csc^{2}x \end{aligned}

Exp and Log ​

Find ddxex\dfrac{d}{dx}e^{x} from first principles.

Solution

ddxex=lim⁡h→0ex+h−exh=lim⁡h→0ex⋅eh−exh=lim⁡h→0ex(eh−1)h=lim⁡h→0ex⋅eh−1h=ex⋅(1)=ex\begin{aligned} \dfrac{d}{dx}e^{x} &= \lim_{h \to 0}\dfrac{e^{x+h}-e^{x}}{h} \\ &= \lim_{h \to 0}\dfrac{e^{x}\cdot e^{h}-e^{x}}{h} \\ &= \lim_{h \to 0}\dfrac{e^{x}(e^{h}-1)}{h} \\ &= \lim_{h \to 0}e^{x}\cdot\dfrac{e^{h}-1}{h} \\ &= e^{x}\cdot(1) \\ &= e^{x} \end{aligned}

Find ddxln⁡x\dfrac{d}{dx}\ln x from first principles.

Solution

ddxln⁡x=lim⁡h→01h[ln⁡(x+h)−ln⁡x]=lim⁡h→01hln⁡(x+hx)=lim⁡h→01hln⁡(1+hx)=lim⁡n→+∞nxln⁡(1+1n)←let hx=1n=lim⁡n→+∞1xln⁡(1+1n)n=1xln⁡e←e=lim⁡n→+∞(1+1n)n=1x\begin{aligned} \dfrac{d}{dx}\ln x &= \lim_{h \to 0}\dfrac{1}{h}\left[\ln(x+h)-\ln x\right] \\ &= \lim_{h \to 0}\dfrac{1}{h}\ln\left(\dfrac{x+h}{x}\right) \\ &= \lim_{h \to 0}\dfrac{1}{h}\ln\left(1+\dfrac{h}{x}\right) \\ &= \lim_{n \to +\infty}\dfrac{n}{x}\ln\left(1+\dfrac{1}{n}\right) && \leftarrow \text{let }\dfrac{h}{x}=\dfrac{1}{n} \\ &= \lim_{n \to +\infty}\dfrac{1}{x}\ln\left(1+\dfrac{1}{n}\right)^{n} \\ &= \dfrac{1}{x}\ln e && \leftarrow e=\lim_{n \to +\infty}\left(1+\dfrac{1}{n}\right)^{n} \\ &= \dfrac{1}{x} \end{aligned}

Find ddxxex\dfrac{d}{dx}xe^{x} from first principles.

Solution

ddxxex=lim⁡h→0(x+h)ex+h−xexh=lim⁡h→0xex+h+hex+h−xexh=lim⁡h→0x(ex+h−ex)+hex+hh=lim⁡h→0[x⋅ex+h−exh+ex+h]=⋯(same steps as ddxex=lim⁡h→0ex+h−exh)=x⋅ex+ex+0=xex+ex\begin{aligned} \dfrac{d}{dx}xe^{x} &= \lim_{h \to 0}\dfrac{(x+h)e^{x+h}-xe^{x}}{h} \\ &= \lim_{h \to 0}\dfrac{xe^{x+h}+he^{x+h}-xe^{x}}{h} \\ &= \lim_{h \to 0}\dfrac{x(e^{x+h}-e^{x})+he^{x+h}}{h} \\ &= \lim_{h \to 0}\left[x\cdot\dfrac{e^{x+h}-e^{x}}{h}+e^{x+h}\right] \\ &= \cdots\left(\text{same steps as }\dfrac{d}{dx}e^{x}=\lim_{h \to 0}\dfrac{e^{x+h}-e^{x}}{h}\right) \\ &= x\cdot e^{x}+e^{x+0} \\ &= xe^{x}+e^{x} \end{aligned}