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satela [25.4K]
3 years ago
10

Derivatives concept: Exercises using the definition of derivatives: (Full development)

Mathematics
1 answer:
krok68 [10]3 years ago
5 0

(A) <em>f(x)</em> = 7 is constant, so <em>f(x</em> + <em>h)</em> = 7, too, which makes <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 0. So <em>f'(x)</em> = 0.

(B) <em>f(x)</em> = 5<em>x</em> + 1   ==>   <em>f(x</em> + <em>h)</em> = 5 (<em>x</em> + <em>h</em>) + 1 = 5<em>x</em> + 5<em>h</em> + 1

==>   <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 5<em>h</em>

Then

\displaystyle f'(x) = \lim_{h\to0}\frac{5h}h = \lim_{h\to0}5 = 5

(C) <em>f(x)</em> = <em>x</em> ² + 3   ==>   <em>f(x</em> + <em>h)</em> = (<em>x</em> + <em>h</em>)² + 3 = <em>x</em> ² + 2<em>xh</em> + <em>h</em> ² + 3

==>   <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 2<em>xh</em> + <em>h</em> ²

\implies\displaystyle f'(x) = \lim_{h\to0}\frac{2xh+h^2}h = \lim_{h\to0}(2x+h) = 2x

(D) <em>f(x)</em> = <em>x</em> ² +<em> </em>4<em>x</em> - 1   ==>   <em>f(x</em> + <em>h)</em> = (<em>x</em> + <em>h</em>)² + 4 (<em>x</em> + <em>h</em>) - 1 = <em>x</em> ² + 2<em>xh</em> + <em>h</em> ² + 4<em>x</em> + 4<em>h</em> - 1

==>   <em>f(x</em> + <em>h)</em> - <em>f(x)</em> = 2<em>xh</em> + <em>h</em> ² + 4<em>h</em>

\implies \displaystyle f'(x) = \lim_{h\to0}\frac{2xh+h^2+4h}h = \lim_{h\to0}(2x+h+4) = 2x+4

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