Question

Find & Prove The Derivative of √tanx=sec²x/2√tanx. -- By Using The First Principal ??

Using First Principal Prove the derivative `√tanx`=`sec^2x/(2√tanx)`

Answer 1

Please see below.

Explanation:

Note: `color(blue)((1)tan(A-B)=(tanA-tanB)/(1+tanAtanB)`

Let , `f(x)=sqrt tanx=>f(t)=sqrt tant`

We know that,

`f'(x)=lim_(t tox)(f(t)-f(x))/(t-x)`

`color(white)(f'(x))=lim_(t tox)(sqrt tant-sqrt tanx)/(t-x)`

`color(white)(f'(x))=lim_(t tox)(sqrt tant-sqrt tanx)/(t-x)xxcolor(green)((sqrt tant+sqrt tanx)/(sqrt tant+sqrt tanx)`

`color(white)(f'(x))=lim_(t tox)1/(sqrt tant+sqrt tanx)*lim_(t tox)(tant-tanx)/(t-x)`

`color(white)(f'(x))`=`1/(sqrt tanx+sqrttanx)lim_(t tox)(tant-tanx)/(t-x)*color(violet)((1+tant tanx)/(1+tant tanx)`

`color(white)(f'(x))`=`1/(2sqrt tanx)lim_(t tox)(color(blue)((tant-tanx)/(1+tant tanx)))/(t-x) (1+tant tanx)tocolor(blue)(Apply(1)`

`color(white)(f'(x))=1/(2sqrttanx)color(red)(lim_((t -x)to0)(tan(t-x)/(t-x)))lim_(t tox)(1+tant tanx)`

`f'(x)=1/(2sqrt tanx)(1)(1+tanxtanx)tocolor(red)([becauselim_(thetato0)(tantheta)/(theta)=1]`

`:.f'(x)=1/(2sqrttanx)(1+tan^2x)`

`=>f'(x)=(sec^2x)/(2sqrt tanx)`