Proof of Various Derivative Properties — Question 9

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Question 9

Prove from first principles that if f(x)=x3,f(x)=x^3, then f′(a)=3a2.f'(a)=3a^2.

Original worksheet page 1: question and worked solution for 7-2-009
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Question 9 - Solution

By definition of the derivative, f′(a)=limh→0f(a+h)−f(a)h.f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.

Substitute f(x)=x3f(x)=x^3: f′(a)=limh→0(a+h)3−a3h.f'(a)=\lim_{h\to 0}\frac{(a+h)^3-a^3}{h}.

Expand the cube: (a+h)3=a3+3a2h+3ah2+h3.(a+h)^3=a^3+3a^2h+3ah^2+h^3.

Substitute into the difference quotient: f′(a)=limh→0a3+3a2h+3ah2+h3−a3h.f'(a)=\lim_{h\to 0}\frac{a^3+3a^2h+3ah^2+h^3-a^3}{h}.

Simplify the numerator: f′(a)=limh→03a2h+3ah2+h3h.f'(a)=\lim_{h\to 0}\frac{3a^2h+3ah^2+h^3}{h}.

Factor out hh: f′(a)=limh→0h(3a2+3ah+h2)h.f'(a)=\lim_{h\to 0}\frac{h(3a^2+3ah+h^2)}{h}.

Cancel hh (for h≠0h\neq 0): f′(a)=limh→0(3a2+3ah+h2).f'(a)=\lim_{h\to 0}(3a^2+3ah+h^2).

Now take the limit: f′(a)=3a2.f'(a)=3a^2.

Thus, using first principles, f′(a)=3a2.\boxed{f'(a)=3a^2}.

Original worksheet page 2: question and worked solution for 7-2-009

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