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Derivative of $x^n$

$$y=x^n$$ $$\text{ln}y=\text{ln}{x}^n$$ $$\text{ln}y=n\text{ln}x$$ $$\frac{1}{y}\frac{\text{d}y}{\text{d}x}=\frac{n}{x}$$ $$\frac{\text{d}y}{\text{d}x}=\frac{ny}{x}$$ $$\frac{\text{d}y}{\text{d}x}=\frac{n{x}^n}{x}$$ $$\frac{\text{d}y}{\text{d}x}=n{x}^{n-1}$$

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Derivation of the product rule

Using differentiation from first principles

$$\frac{d}{dx}(uv)=\underset{h\to 0}{Lim}\left(\frac{u(x+h)v(x+h)-u\left(x\right)v\left(x\right)}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{u(x+h)v(x+h)-u(x+h)v\left(x\right)+u(x+h)v\left(x\right)-u\left(x\right)v\left(x\right)}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{u(x+h)\left(v\right(x+h)-v(x\left)\right)+v\left(x\right)\left(u\right(x+h)-u(x\left)\right)}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{u(x+h)\left(v\right(x+h)-v(x\left)\right)}{h}\right)+\underset{h\to 0}{Lim}\left(\frac{v\left(x\right)\left(u\right(x+h)-u(x\left)\right)}{h}\right)$$ $$=u(x)\underset{h\to 0}{Lim}\left(\frac{v(x+h)-v\left(x\right)}{h}\right)+v(x)\underset{h\to 0}{Lim}\left(\frac{u(x+h)-u\left(x\right)}{h}\right)$$ $$=u(x)\frac{d}{dx}v(x)+v(x)\frac{d}{dx}u(x)$$ $$=\frac{du}{dx}v+u\frac{du}{dx}$$

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Proof of $\frac{d}{dx}(lnx)=\frac{1}{x}$

$$y=lnx$$ $${\mathrm{e}}^y=x$$ $${\mathrm{e}}^y\frac{dy}{dx}=1$$ $$\frac{dy}{dx}=\frac{1}{{\mathrm{e}}^y}$$ $$\frac{d}{dx}(lnx)=\frac{1}{x}$$

... as required.

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Derivaitve of $cosx$

Using differentiation from first principles

$$\frac{d}{dx}cosx=\underset{h\to 0}{Lim}\left(\frac{cos(x+h)-cosx}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{cosxcosh-sinxsinh-cosx}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{(1-\frac{h^2}{2})cosx-hsinx-cosx}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{cosx-\frac{h^2}{2}cosx-hsinx-cosx}{h}\right)$$ $$=\underset{h\to 0}{Lim}(-\frac{h}{2}cosx-sinx)$$ $$=-sinx$$

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Derivaitve of $sinx$

Using differentiation from first principles

$$\frac{d}{dx}sinx=\underset{h\to 0}{Lim}\left(\frac{sin(x+h)-sinx}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{sinxcosh+sinhcosx-sinx}{h}\right)$$ $$=\underset{h\to 0}{Lim}\left(\frac{sinx+hcosx-sinx}{h}\right)$$ $$=\underset{h\to 0}{Lim}(cosx)$$ $$=cosx$$

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Derivative of $tanx$

$$y=tanx$$ $$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{sinx}{cosx}\right)$$ $$=\frac{\frac{dsinx}{dx}cosx-sinx\frac{dcosx}{dx}}{{cos}^{2}x}$$ $$=\frac{{cos}^{2}x+{sin}^{2}x}{{cos}^{2}x}$$ $$=1+{tan}^{2}x$$ $$={sec}^{2}x$$