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