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How do you find the derivative of #f(x)=1-x^2# using the limit process?

Answer 1

Isaiah Allen

#d/(dx) (1-x^2) = -2x#

By definition the derivative is:

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

Evaluate the limit in the case: #f(x) = 1-x^2#:

#d/(dx) (1-x^2) = lim_(h->0) ((1-(x+h)^2-(1-x^2))/h)#

#:.d/(dx) (1-x^2) = lim_(h->0) (cancel(1)-cancel(x^2)-2hx-h^2cancel(-1)+cancel(x^2))/h#

#:.d/(dx) (1-x^2) = lim_(h->0) (-2hx-h^2)/h = lim_(h->0) (cancel(h)(-2x-h))/cancel(h)#

#:.d/(dx) (1-x^2) = lim_(h->0) (-2x-h) = -2x#

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Answer 2

Emilia Aguilar

To find the derivative of ( f(x) = 1 - x^2 ) using the limit process, you apply the definition of the derivative:

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

Substitute the function ( f(x) = 1 - x^2 ) into the definition:

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

Expand and simplify:

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

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

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

Now, take the limit as ( h ) approaches 0:

[ f'(x) = -2x ]

Therefore, the derivative of ( f(x) = 1 - x^2 ) with respect to ( x ) is ( f'(x) = -2x ).

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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.