Question

Differentiate using first principle : e^ [under root(tan x)]

Answer 1

Answer:

Explanation:

Using first principle, derivative of  : e [under root(tan x)] is

e^√tanx sec²x/2√tanx

f ( x ) =

By first principle,

f'(x) = lim h->0  (f(x +h) - f (x)) /h ==>

f'(x) = lim h->0 e^√tan(x+h) - e^√tanx / h =

Taking e^√tanx outside

f'(x) =  (e^√tanx ) lim h->0  ( e^√tan(x+h) -  √tanx -  1 )/ h

Multiplying and dividing by √tan(x+h) - √tanx and √tan(x+h) + √tanx

f'(x) = (e^√tanx ) limh->0{ (e^(√tan(x+h) -  √tanx )-  1 )/√tan(x+h) - √tanx } x { √tan(x+h) - √tanx  / √tan(x+h) + √tanx} x { √tan(x+h) + √tanx / h }

We know lim x-> 0 (e^x - 1)/x = 1

Therefore,

limh->0{ (e^(√tan(x+h) -  √tanx )-  1 )/√tan(x+h) - √tanx } = 1

Therefore,

f'(x) =  (e^√tanx ) limh->0 { (tan(x+h) - tanx)/h(√tan(x+h) + √tanx)

Also tan(A - B) =  tanA - tanB / ( 1 +tan(A) tan(B))

tan(x+h) - tanx = tan (x+h - x) (1 + tan(x +h)tan x)

lim x - > 0 tanx/x = 1

Therefore,

f'(x) =  (e^√tanx ) limh->0 { tan (x+h - x) (1 + tan(x +h)tan x) /h(√tan(x+h) + √tanx)}

Applying limit,

f'(x)  =   (e^√tanx )  ( 1 + tan²x)/ 2√tanx =

f'(x) =  e^√tanx sec²x/2√tanx