How do you find f'(x) using the definition of a derivative for f(x)=ex?

The definition of the first derivative isf′(x)=limx→0f(x+h)−f(x)hhence for f(x)=ex we have thatf′(x)=limh→0ex+h−exh=limh→0ex⋅(eh−1h)=ex⋅limh→0(eh−1h)=exRemarkslimh→0(eh−1h)=1

The derivative of f(x) with respect to x is defined as:limh→0(ex+h−exh)limh→0(ex+h−exh)=ex=ex⇒1

The definition of the first derivative isf′(x)=limx→0f(x+h)−f(x)hhence for f(x)=ex we have thatf′(x)=limh→0ex+h−exh=limh→0ex⋅(eh−1h)=ex⋅limh→0(eh−1h)=exRemarkslimh→0(eh−1h)=1

The derivative of f(x) with respect to x is defined as:limh→0(ex+h−exh)limh→0(ex+h−exh)=ex=ex⇒1

Find out the answer to "How do you find f'(x) using the definition..."

High School
Calculus and Analysis
Calculus 1
Derivatives

Ekwadorkadba3

Answered question

2022-04-16

How do you find f'(x) using the definition of a derivative for

f (x) = e^{x}

Answer & Explanation

Dallelopeelvep2yc

Beginner2022-04-17Added 15 answers

The definition of the first derivative is

f^{'} (x) = lim_{x \to 0} \frac{f (x + h) - f (x)}{h}

hence for

f (x) = e^{x}

we have that

f^{'} (x) = lim_{h \to 0} \frac{e^{x + h} - e^{x}}{h} = lim_{h \to 0} e^{x} \cdot (\frac{e^{h} - 1}{h}) = e^{x} \cdot lim_{h \to 0} (\frac{e^{h} - 1}{h}) = e^{x}

Remarks

lim_{h \to 0} (\frac{e^{h} - 1}{h}) = 1

mortaraskxnl

Beginner2022-04-18Added 12 answers

The derivative of f(x) with respect to x is defined as:

lim_{h \to 0} (\frac{e^{x + h} - e^{x}}{h})

lim_{h \to 0} (\frac{e^{x + h} - e^{x}}{h}) = e^{x}

= e^{x} \Rightarrow 1

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