# How do you find f'(x) using the definition of a derivative for #f(x)= (4+x) / (1-4x)#?

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To find ( f'(x) ) using the definition of a derivative for ( f(x) = \frac{{4+x}}{{1-4x}} ), we apply the definition of the derivative:

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

Substitute the function ( f(x) = \frac{{4+x}}{{1-4x}} ) into the definition:

[ f'(x) = \lim_{{h \to 0}} \frac{{\frac{{4+(x+h)}}{{1-4(x+h)}} - \frac{{4+x}}{{1-4x}}}{{h}} ]

Simplify the expression:

[ f'(x) = \lim_{{h \to 0}} \frac{{\frac{{4+x+h}}{{1-4x-4h}} - \frac{{4+x}}{{1-4x}}}{{h}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{(4+x+h)(1-4x) - (4+x)(1-4x-4h)}}{{h(1-4x)(1-4x-4h)}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{(4+x+h)(1-4x) - (4+x)(1-4x-4h)}}{{h(1-4x)(1-4x-4h)}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{4+ x + h -16x -4xh - 4 + 4x + 4x^2 + 4h - 4x - 4x^2 - 4h}}{{h(1-4x)(1-4x-4h)}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{x -16x - 4xh + 4x + 4h - 4}}{{h(1-4x)(1-4x-4h)}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{-15x - 4xh + 4x + 4h - 4}}{{h(1-4x)(1-4x-4h)}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{-15x + 4x - 4 + h(-4 -4x)}}{{h(1-4x)(1-4x-4h)}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{-11x - 4}}{{(1-4x)(1-4x-4h)}} ]

Now, we can substitute ( h = 0 ) into the expression:

[ f'(x) = \frac{{-11x - 4}}{{(1-4x)(1-4x)}} ]

[ f'(x) = \frac{{-11x - 4}}{{(1-4x)^2}} ]

So, ( f'(x) = \frac{{-11x - 4}}{{(1-4x)^2}} ).

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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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