Given-f-x-5-x-find-f-x-by-using-limit-first-principal-derivative-

Question Number 116930 by bobhans last updated on 08/Oct/20

Given f (x) = 5^{\sqrt{x}} find f^{'} (x) by using limit

(first principal derivative)

Commented by bobhans last updated on 08/Oct/20

Answered by 1549442205PVT last updated on 09/Oct/20

First, we prove that \lim_{x \to 0} \frac{a^{x} - 1}{x} = lna .

by using the definition of limit

We can assume x > 0 (if x < 0 put x = - t)

Suppose ϵ > 0 be an arbitrarily small number

so that ∣ \frac{a^{x} - 1}{x} - lna ∣< ϵ . Then choosing

δ = \frac{\ln [{(- ϵ x + xlna + 1)}^{- 1}]}{lna} > 0 we have

∣ x ∣< δ \Leftrightarrow - \frac{\ln [{(- ϵ x + xlna + 1)}^{- 1}]}{lna} < x < \frac{\ln [{(- ϵ x + xlna + 1)}^{- 1}]}{lna}

\Rightarrow - \ln [{(- ϵ x + xlna + 1)}^{- 1}] < xlna < \ln [{(- ϵ x + xlna + 1)}^{- 1}]

\Rightarrow \ln (- ϵ x + xlna + 1) < {lna}^{x} < \ln [{(- ϵ x + xlna + 1)}^{- 1}]

\Rightarrow - ϵ x + xlna + 1 < a^{x} < {(- ϵ x + xlna + 1)}^{- 1}

\Rightarrow - ϵ x < a^{x} - 1 - xlna < ϵ x

(Since {(- ϵ x + xlna + 1)}^{- 1} = \frac{1}{- ϵ x + xlna + 1}

< - ϵ x + xlna + 1 < ϵ x + xlna + 1)

\Rightarrow - ϵ < \frac{a^{x} - 1 - xlna}{x} < ϵ \Rightarrow∣ \frac{a^{x} - 1}{x} - lna ∣< ϵ

By the definition of the limit of a

function we have :

\lim_{x \to 0} \frac{a^{x} - 1}{x} = lna (Q . E . D) (1)

Now by the definition of derivative of a

function we have

f^{'} (x) = \lim {[f (x + Δ x) - f (x)] / Δ x} . Hence,

{(5^{\sqrt{x}})}^{'} = \lim_{Δ u \to 0} \frac{5^{\sqrt{x + Δ x}} - 5^{\sqrt{x}}}{Δ x}

\underset{Δ x \to 0}{= \lim} \frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) . \frac{Δ x}{\sqrt{x + Δ x} - \sqrt{x}}}

= \lim \frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) . \frac{Δ x (\sqrt{x + Δ x} + \sqrt{x})}{\sqrt{x + Δ x} - \sqrt{x}) (\sqrt{x + Δ x} + \sqrt{x})}}

= \lim \frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) . \frac{Δ x (\sqrt{x + Δ x} + \sqrt{x})}{Δ x}}

= \lim_{Δ x \to 0} (\frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) .}) . \lim_{Δ x \to 0} \frac{1}{\sqrt{x + Δ x} + \sqrt{x}}

= 5^{\sqrt{x}} \ln 5 . \frac{1}{2 \sqrt{x}}

since \lim_{Δ x \to 0} \frac{(5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) .} = \ln 5 (by (1))

Thus, finally we obtained

{(a^{\sqrt{x}})}^{'} = \frac{5^{\sqrt{x}} \ln 5}{2 \sqrt{x}} is proved by limit

Commented by bemath last updated on 08/Oct/20

sir why not \frac{5^{\sqrt{x}}}{2 \sqrt{x}} . \ln (5) ?

Commented by bemath last updated on 08/Oct/20

(5^{\sqrt{x + Δ x}} - 5^{\sqrt{x}}) \times (5^{\sqrt{x + Δ x}} + 5^{\sqrt{x}})

\neq 5^{x + Δ x} - 5^{x} sir

Commented by bobhans last updated on 08/Oct/20

sir pvt, your answer not correct sir

Commented by 1549442205PVT last updated on 08/Oct/20

I had a mistake and i am correcting it

and now it is corrected completely

Please, check help me . Thank Sir

Commented by bemath last updated on 08/Oct/20

thank you

Commented by bobhans last updated on 08/Oct/20

sir pvt, thank you

Commented by 1549442205PVT last updated on 09/Oct/20

Thank Sir . You are welcome .

Answered by TANMAY PANACEA last updated on 08/Oct/20

\frac{d y}{d x} = \lim_{h \to 0} \frac{5^{\sqrt{x + h}} - 5^{\sqrt{x}}}{h}

= \lim_{h \to 0} 5^{\sqrt{x}} \times (\frac{5^{\sqrt{x + h} - \sqrt{x}} - 1}{h})

= 5^{\sqrt{x}} \times \lim_{t \to 0} (\frac{5^{t} - 1}{t}) \times \lim_{h \to 0} (\frac{\sqrt{x + h} - \sqrt{x}}{h}) [t = \sqrt{x + h} - \sqrt{x}]

= 5^{\sqrt{x}} \times \lim_{t \to 0} (\frac{e^{t l n 5} - 1}{t l n 5}) \times l n 5 \times \lim_{h \to 0} (\frac{h}{h} \times \frac{1}{\sqrt{x + h} + \sqrt{x}})

= 5^{\sqrt{x}} \times 1 \times l n 5 \times \frac{1}{2 \sqrt{x}} = 5^{\sqrt{x}} \times l n 5 \times \frac{1}{2 \sqrt{x}}