How do you use the definition of a derivative to find f' given #f(x)=1/x^2# at x=1?

Answer 1

#-2#

#f(x) = 1/x^2#

The derivative is defined as:

#lim_(h->0)(f(x+h)-f(x))/h#

Which in our case would be:

#=lim_(h->0)(1/(x+h)^2-1/x^2)/h=lim_(h->0)(((x^2-(x+h)^2)/(x^2(x+h)^2)))/h#
#=lim_(h->0)(x^2-(x+h)^2)/(x^2(x+h)^2h)#
Substitute #x=1#:
#=lim_(h->0)(1^2-(1+h)^2)/(1^2(1+h)^2h)#
#=lim_(h->0)(1-(1+h)^2)/((1+h)^2h)=lim_(h->0)(1-1-2h-h^2)/((1+h)^2h)#
#lim_(h->0)(-2h-h^2)/((1+h)^2h)=lim_(h->0)(-(2+h)h)/((1+h)^2h)#
#lim_(h->0)(-(2+h))/((1+h)^2)#

Evaluate the limit:

#=-2/1=-2#
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Answer 2

To use the definition of a derivative to find (f'(x)) given (f(x) = \frac{1}{x^2}) at (x = 1), we use the definition of the derivative:

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

Substitute (f(x) = \frac{1}{x^2}):

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

Now, plug in (x = 1):

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

Simplify:

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

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

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

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

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

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

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

[f'(1) = \frac{-2}{(1)^2}]

[f'(1) = -2]

Therefore, (f'(1) = -2).

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Answer 3

To use the definition of a derivative to find ( f'(x) ) given ( f(x) = \frac{1}{x^2} ) at ( x = 1 ), we'll use the definition:

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

First, find ( f(x+h) ):

[ f(x+h) = \frac{1}{(x+h)^2} ]

Now, substitute ( f(x+h) ) and ( f(x) ) into the definition:

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

Next, simplify the expression:

[ f'(x) = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{h \cdot x^2 \cdot (x+h)^2} ]

[ f'(x) = \lim_{h \to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{h \cdot x^2 \cdot (x^2 + 2xh + h^2)} ]

[ f'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h \cdot x^2 \cdot (x^2 + 2xh + h^2)} ]

Now, cancel out ( h ) from the numerator and denominator:

[ f'(x) = \lim_{h \to 0} \frac{-2x - h}{x^2 \cdot (x^2 + 2xh + h^2)} ]

Finally, evaluate the limit as ( h ) approaches 0:

[ f'(x) = \frac{-2x}{x^2 \cdot (x^2)} ]

[ f'(x) = \frac{-2x}{x^4} ]

Now, substitute ( x = 1 ) into the expression to find ( f'(1) ):

[ f'(1) = \frac{-2 \cdot 1}{1^4} ]

[ f'(1) = -2 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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