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Zorluk: KolayDifferentiation from First Principles

Using differentiation from first principles, evaluate the value of the derivative of the function f(x)=2x2+3x1f(x) = 2x^2 + 3x - 1 at the point where x=1x = 1.

Cevap: 7

Cevap

The value of the derivative at x=1x = 1 is 7.
Using first principles, the derivative is defined as f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. Evaluating at x=1x = 1, we find f(1)=4f(1) = 4 and f(1+h)=4+7h+2h2f(1+h) = 4 + 7h + 2h^2. Subtracting f(1)f(1) leaves 7h+2h27h + 2h^2, and dividing by hh yields 7+2h7 + 2h. Taking the limit as h0h \to 0 yields the final result of 7.

Adım Adım Çözüm

1
Calculate the value of the function at x=1x = 1
f(1)=2(1)2+3(1)1=4f(1) = 2(1)^2 + 3(1) - 1 = 4
This establishes the base value needed for the difference quotient.
2
Expand and simplify f(1+h)f(1+h)
f(1+h)=2(1+h)2+3(1+h)1=4+7h+2h2f(1+h) = 2(1+h)^2 + 3(1+h) - 1 = 4 + 7h + 2h^2
This gives the value of the function at the incremented point x+hx + h.
3
Form and simplify the difference quotient f(1+h)f(1)h\frac{f(1+h) - f(1)}{h}
4+7h+2h24h=7+2h\frac{4 + 7h + 2h^2 - 4}{h} = 7 + 2h
Dividing by hh eliminates the indeterminate form before taking the limit.
4
Evaluate the limit as h0h \to 0
f(1)=limh0(7+2h)=7f'(1) = \lim_{h \to 0} (7 + 2h) = 7
Taking h=0h = 0 in the simplified quotient gives the exact rate of change at x=1x = 1.

Anahtar Kavram

Differentiation from First Principles
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