Solved: "Using the limit definition, how do you differentiate..."

ngihlungeqtr

Answered question

2022-04-21

Using the limit definition, how do you differentiate

f (x) = x^{2} + 3 x + 1

Sarai Dorsey

Beginner2022-04-22Added 16 answers

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

In this case

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

= lim_{h \to 0} \frac{{(x + h)}^{2} + 3 (x + h) + 1 - x^{2} - 3 x - 1}{h}

= lim_{h \to 0} \frac{x^{2} + 2 x h + h^{2} + 3 x + 3 h + 1 - x^{2} - 3 x - 1}{h}

= lim_{h \to 0} \frac{2 x h + h^{2} + 3 h}{h}

= lim_{h \to 0} 2 x + h + 3 = 2 x + 3

timbreoizy

Beginner2022-04-23Added 15 answers

The definition of the derivative is:

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

Our function f(x) equals

x^{2} + 3 x + 1

. Applying it to the definition, we have:

f^{'} (x) = lim_{h \to 0} \frac{{(x + h)}^{2} + 3 (x + h) + 1 - (x^{2} + 3 x + 1)}{h}

And doing some algebra to finish off:

f^{'} (x) = lim_{h \to 0} \frac{x^{2} + 2 x h + h^{2} + 3 x + 3 h + 1 - x^{2} - 3 x - 1}{h}

f^{'} (x) = lim_{h \to 0} \frac{2 x h + h^{2} + 3 h}{h}

f^{'} (x) = lim_{h \to 0} \frac{h (2 x + h + 3)}{h}

f^{'} (x) = lim_{h \to 0} 2 x + h + 3

f'(x)=2x+(0)+3
f'(x)=2x+3

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