Differentiate y= tanx using first principle

Jete Igbodo
2022-07-19
Answered

Differentiate y= tanx using first principle

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Jeffrey Jordon

Answered 2022-11-07
Author has **2581** answers

$1+{\mathrm{tan}}^{2}\left(x\right)={\mathrm{sec}}^{2}\left(x\right)$

$\mathrm{tan}\left(a+b\right)=\frac{\mathrm{tan}\left(a\right)+\mathrm{tan}\left(b\right)}{1-\mathrm{tan}\left(a\right)\mathrm{tan}\left(b\right)}$

and

$\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{tan}\left(x\right)}{x}=1$

$\text{Let}f(x)=\mathrm{tan}x$

${f}^{\prime}(x)={\displaystyle \underset{a\to 0}{lim}{\displaystyle \frac{f(x+a)-f(x)}{a}}}$

$={\displaystyle \underset{a\to 0}{lim}{\displaystyle \frac{\mathrm{tan}(x+a)-\mathrm{tan}x}{a}}}$

$={\displaystyle \underset{a\to 0}{lim}{\displaystyle \frac{\mathrm{tan}x+\mathrm{tan}a-\mathrm{tan}x+{\mathrm{tan}}^{2}x\mathrm{tan}a}{a(1-\mathrm{tan}x\mathrm{tan}a)}}}$

$={\displaystyle \underset{a\to 0}{lim}{\displaystyle \frac{\mathrm{tan}a(1+{\mathrm{tan}}^{2}x)}{a(1-\mathrm{tan}x\mathrm{tan}a)}}}$

$={\displaystyle \underset{a\to 0}{lim}{\displaystyle \frac{\mathrm{tan}a{\mathrm{sec}}^{2}a}{a(1-\mathrm{tan}x\mathrm{tan}a)}}}$

Therefore

$f\text{'}\left(x\right)={\mathrm{sec}}^{2}\left(x\right)$

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