Calculating Derivatives 2

\cos(x)

\ln(x)

\begin{align*} f(x)&=\cos(x) \\\\ f'(x)&=\lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \\\\ &=\lim_{h \to 0}\frac{\c1{\cos(x+h)} - \c1{\cos(x)}}{h} \\\\ &=\lim_{h \to 0}\frac{\c1{\cos(x)\cos(h) - \sin(x)\sin(h)} - \cos(x)}{h} \\\\ &=\lim_{h \to 0}\frac{\c1{\cos(x)[\cos(h) - 1 ]} - \sin(x)\sin(h)}{h} \\\\ &=\c1{\lim_{h \to 0}}\frac{\cos(x)[\cos(h) - 1 ]}{h} - \c1{\lim_{h \to 0}}\frac{\sin(x)\sin(h)}{h} \\\\ &=\c1{\cos(x)}\lim_{h \to 0}\frac{\cos(h) - 1 }{h} - \c1{\sin(x)}\lim_{h \to 0}\frac{\sin(h)}{h} \\\\ &=\cos(x)\c1{[0]} - \sin(x)\c1{[1]} \\\\ &=-\sin(x) \end{align*}

\begin{align*} f(x) &= \ln(x) \\\\ f'(x)&=\lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \\\\ &=\lim_{h \to 0}\frac{\c1{\ln(x+h)} - \c1{\ln(x)}}{h} \\\\ &=\lim_{h \to 0}\frac{\c1{\ln(\frac{x+h}{x})}}{h} \\\\ &=\lim_{h \to 0}\ln \left(\frac{x+h}{x} \right)^{\c1{1/h}} \\\\ &=\lim_{h \to 0}\ln \left(\c1{1+\frac{h}{x}} \right)^{1/h} \\\\ t &= \frac{h}{x} \\\\ f'(x)&=\lim_{t \to 0}\ln(1+\c1{t})^{\c1{(1/x)(1/t)}} \\\\ &=\lim_{t \to 0}\frac{\ln((1+t)^{1/t})}{\c1{x}} \\\\ &=\c1{\frac{1}{x}}\lim_{t \to 0}\ln((1+t)^{1/t}) \\\\ &=\frac{1}{x}\c1{\ln}(\,\c1{\lim_{t \to 0}}(1+t)^{1/t}) \\\\ \text{Note } e &= \lim_{t \to 0}(1+t)^{1/t} \\\\ f'(x)&=\frac{1}{x}\c1{\ln(e)} \\\\ &=\frac{1}{x} \end{align*}