Answered

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Using the limit definition of a derivative, what is:5x^2 + 2x

Sagot :

Let:

[tex]f(x)=5x^2+2x[/tex]

Using the limit definition of a derivative:

[tex]f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]

so:

[tex]\begin{gathered} \lim _{h\to0}\frac{5(x+h)^2+2(x+h)-5x^2-2x}{h} \\ \lim _{h\to0}\frac{5x^2+10xh+5h^2+2x+2h-5x^2-2x}{h} \\ \end{gathered}[/tex]

Add like terms:

[tex]\lim _{h\to0}\frac{10xh+5h^2+2h}{h}[/tex]

Divide the numerator and the denominator by h:

[tex]\begin{gathered} \lim _{h\to0}10x+5h+2=10x+5(0)+2=10x+2 \\ so\colon \\ f^{\prime}(x)=10x+2 \end{gathered}[/tex]

Answer:

f'(x) = 10x + 2

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