# How do I differentiate y=tan(x) from the first principle?

Differentiate y= tanx using first principle

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Jeffrey Jordon

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

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

and

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

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

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

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

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

Therefore

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