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Derivative of x^4 from the limit definition

Let f(x)=x4f(x) = x^4. Using the limit definition of the derivative, f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, find f′(x)f'(x). Show the expansion of (x+h)4(x+h)^4, simplify the difference quotient, and evaluate the limit as h→0h \to 0.

Hint

Expand (x+h)4(x+h)^4 using the binomial theorem, subtract x4x^4, factor out hh from the numerator, then cancel the hh in the denominator before taking the limit.

Step-by-step solution

Start by expanding (x+h)4=x4+4x3h+6x2h2+4xh3+h4(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4. Substituting into the difference quotient gives f(x+h)−f(x)h=x4+4x3h+6x2h2+4xh3+h4−x4h\dfrac{f(x+h)-f(x)}{h} = \dfrac{x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 - x^4}{h}. The x4x^4 terms cancel, leaving 4x3h+6x2h2+4xh3+h4h\dfrac{4x^3h + 6x^2h^2 + 4xh^3 + h^4}{h}.

Factor hh out of the numerator to get h(4x3+6x2h+4xh2+h3)h\dfrac{h(4x^3 + 6x^2h + 4xh^2 + h^3)}{h}, and cancel the common factor of hh (valid since h≠0h \neq 0 in the limit) to obtain 4x3+6x2h+4xh2+h34x^3 + 6x^2h + 4xh^2 + h^3. Taking the limit as h→0h \to 0, every term containing hh vanishes, leaving f′(x)=4x3f'(x) = 4x^3.

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