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Zorluk: KolayDifferentiation from First Principles

Using differentiation from first principles, what is the value of the derivative of the function f(x)=x2+2xf(x) = x^2 + 2x at the point where x=3x = 3?

Cevap: 8

Cevap

The derivative of f(x)=x2+2xf(x) = x^2 + 2x evaluated at x=3x = 3 is 8.
Applying first principles to f(x)=x2+2xf(x) = x^2 + 2x yields f(x)=limh0(x+h)2+2(x+h)(x2+2x)h=limh0(2x+h+2)=2x+2f'(x) = \lim_{h \to 0} \frac{(x+h)^2 + 2(x+h) - (x^2 + 2x)}{h} = \lim_{h \to 0} (2x + h + 2) = 2x + 2. Evaluating this derivative at x=3x = 3 gives 2(3)+2=82(3) + 2 = 8.

Adım Adım Çözüm

1
Apply the definition of differentiation from first principles
f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
First principles uses the limit of the difference quotient to compute the instantaneous rate of change.
2
Substitute f(x)=x2+2xf(x) = x^2 + 2x into the difference quotient
[(x+h)2+2(x+h)](x2+2x)h=2xh+h2+2hh\frac{[(x+h)^2 + 2(x+h)] - (x^2 + 2x)}{h} = \frac{2xh + h^2 + 2h}{h}
Expanding terms allows cancellation of non-hh terms in the numerator.
3
Simplify the fraction by dividing by hh
2x+h+22x + h + 2
Dividing out hh removes the indeterminate form 00\frac{0}{0}.
4
Evaluate the limit as h0h \to 0 and substitute x=3x = 3
f(3)=2(3)+2=8f'(3) = 2(3) + 2 = 8
Setting h=0h = 0 yields the derivative function f(x)=2x+2f'(x) = 2x + 2, which evaluates to 8 at x=3x = 3.

Anahtar Kavram

Differentiation from First Principles
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