Given f(x)=5^(√x) find f ′(x) by using limit (first principal derivative)


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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
	$First,we prove that lim _(x→0) ((a^x −1)/x)=lna. by using the definition of limit We can assume x>0(if x<0 put x=−t) Suppose ε>0 be an arbitrarily small number so that ∣((a^x −1)/x)−lna∣<ε.Then choosing δ=((ln[(−εx+xlna+1)^(−1) ])/(lna))>0 we have ∣x∣<δ⇔−((ln[(−εx+xlna+1)^(−1) ])/(lna))<x<((ln[(−εx+xlna+1)^(−1) ])/(lna)) ⇒−ln[(−εx+xlna+1)^(−1) ]<xlna<ln[(−εx+xlna+1)^(−1) ] ⇒ln(−εx+xlna+1)<lna^x <ln[(−εx+xlna+1)^(−1) ] ⇒−εx+xlna+1<a^x <(−εx+xlna+1)^(−1) ⇒−εx<a^x −1−xlna<εx (Since (−εx+xlna+1)^(−1) =(1/(−εx+xlna+1)) <−εx+xlna+1<εx+xlna+1) ⇒−ε<((a^x −1−xlna)/x)<ε⇒∣((a^x −1)/x)−lna∣<ε By the definition of the limit of a function we have: lim_(x→0) ((a^x −1)/x)=lna (Q.E.D)(1) Now by the definition of derivative of a function we have f ′(x)=lim{[f(x+Δx)−f(x)]/Δx}.Hence, (5^(√x) )′=lim_(Δu→0) ((5^(√(x+Δx)) −5^(√x) )/( Δx)) =lim_(Δx→0) ((5^(√x) (5^((√(x+Δx))−(√x)) −1))/( ((√(x+Δx))−(√x)).((Δx)/( (√(x+Δx))−(√x))))) =lim((5^(√x) (5^((√(x+Δx))−(√x)) −1))/( ((√(x+Δx))−(√x)).((Δx((√(x+Δx))+(√x)))/( (√(x+Δx))−(√x))((√(x+Δx))+(√x)))))) =lim((5^(√x) (5^((√(x+Δx))−(√x)) −1))/( ((√(x+Δx))−(√x)).((Δx((√(x+Δx))+(√x)))/( Δx)))) =lim_(Δx→0) (((5^(√x) (5^((√(x+Δx))−(√x)) −1))/( ((√(x+Δx))−(√x)).))).lim_(Δx→0) (1/( (√(x+Δx))+(√x))) =5^(√x) ln5.(1/(2(√x))) since lim_(Δx→0) (((5^((√(x+Δx))−(√x)) −1))/( ((√(x+Δx))−(√x)).))=ln5(by (1)) Thus,finally we obtained (a^(√x) )′=((5^(√x) ln5)/(2(√x))) is proved by limit$
	$First, we prove that \lim_{x \to 0} \frac{a^{x} - 1}{x} = lna .$ $by using the definition of limit$ $We can assume x > 0 (if x < 0 put x = - t)$ $Suppose ϵ > 0 be an arbitrarily small number$ $so that ∣ \frac{a^{x} - 1}{x} - lna ∣< ϵ . Then choosing$ $δ = \frac{\ln [{(- ϵ x + xlna + 1)}^{- 1}]}{lna} > 0 we have$ $∣ x ∣< δ \Leftrightarrow - \frac{\ln [{(- ϵ x + xlna + 1)}^{- 1}]}{lna} < x < \frac{\ln [{(- ϵ x + xlna + 1)}^{- 1}]}{lna}$ $\Rightarrow - \ln [{(- ϵ x + xlna + 1)}^{- 1}] < xlna < \ln [{(- ϵ x + xlna + 1)}^{- 1}]$ $\Rightarrow \ln (- ϵ x + xlna + 1) < {lna}^{x} < \ln [{(- ϵ x + xlna + 1)}^{- 1}]$ $\Rightarrow - ϵ x + xlna + 1 < a^{x} < {(- ϵ x + xlna + 1)}^{- 1}$ $\Rightarrow - ϵ x < a^{x} - 1 - xlna < ϵ x$ $(Since {(- ϵ x + xlna + 1)}^{- 1} = \frac{1}{- ϵ x + xlna + 1}$ $< - ϵ x + xlna + 1 < ϵ x + xlna + 1)$ $\Rightarrow - ϵ < \frac{a^{x} - 1 - xlna}{x} < ϵ \Rightarrow∣ \frac{a^{x} - 1}{x} - lna ∣< ϵ$ $By the definition of the limit of a$ $function we have :$ $\lim_{x \to 0} \frac{a^{x} - 1}{x} = lna (Q . E . D) (1)$ $Now by the definition of derivative of a$ $function we have$ $f^{'} (x) = \lim {[f (x + Δ x) - f (x)] / Δ x} . Hence,$ ${(5^{\sqrt{x}})}^{'} = \lim_{Δ u \to 0} \frac{5^{\sqrt{x + Δ x}} - 5^{\sqrt{x}}}{Δ x}$ $\underset{Δ x \to 0}{= \lim} \frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) . \frac{Δ x}{\sqrt{x + Δ x} - \sqrt{x}}}$ $= \lim \frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) . \frac{Δ x (\sqrt{x + Δ x} + \sqrt{x})}{\sqrt{x + Δ x} - \sqrt{x}) (\sqrt{x + Δ x} + \sqrt{x})}}$ $= \lim \frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) . \frac{Δ x (\sqrt{x + Δ x} + \sqrt{x})}{Δ x}}$ $= \lim_{Δ x \to 0} (\frac{5^{\sqrt{x}} (5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) .}) . \lim_{Δ x \to 0} \frac{1}{\sqrt{x + Δ x} + \sqrt{x}}$ $= 5^{\sqrt{x}} \ln 5 . \frac{1}{2 \sqrt{x}}$ $since \lim_{Δ x \to 0} \frac{(5^{\sqrt{x + Δ x} - \sqrt{x}} - 1)}{(\sqrt{x + Δ x} - \sqrt{x}) .} = \ln 5 (by (1))$ $Thus, finally we obtained$ ${(a^{\sqrt{x}})}^{'} = \frac{5^{\sqrt{x}} \ln 5}{2 \sqrt{x}} is proved by limit$
	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

	$\frac{d y}{d x} = \lim_{h \to 0} \frac{5^{\sqrt{x + h}} - 5^{\sqrt{x}}}{h}$ $= \lim_{h \to 0} 5^{\sqrt{x}} \times (\frac{5^{\sqrt{x + h} - \sqrt{x}} - 1}{h})$ $= 5^{\sqrt{x}} \times \lim_{t \to 0} (\frac{5^{t} - 1}{t}) \times \lim_{h \to 0} (\frac{\sqrt{x + h} - \sqrt{x}}{h}) [t = \sqrt{x + h} - \sqrt{x}]$ $= 5^{\sqrt{x}} \times \lim_{t \to 0} (\frac{e^{t l n 5} - 1}{t l n 5}) \times l n 5 \times \lim_{h \to 0} (\frac{h}{h} \times \frac{1}{\sqrt{x + h} + \sqrt{x}})$ $= 5^{\sqrt{x}} \times 1 \times l n 5 \times \frac{1}{2 \sqrt{x}} = 5^{\sqrt{x}} \times l n 5 \times \frac{1}{2 \sqrt{x}}$
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