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Derivative of 2/x^2 via limit definition

Consider the function f(x)=2x2f(x) = \dfrac{2}{x^2}.

Use 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}, to find f′(x)f'(x) for f(x)=2x2f(x) = \dfrac{2}{x^2}. Show the algebraic steps that lead to your final simplified expression.

Hint

Write f(x+h)−f(x)=2(x+h)2−2x2f(x+h) - f(x) = \dfrac{2}{(x+h)^2} - \dfrac{2}{x^2} and combine into a single fraction with common denominator x2(x+h)2x^2(x+h)^2 before dividing by hh.

Step-by-step solution

Start by forming the difference quotient: f(x+h)−f(x)h=1h(2(x+h)2−2x2).\frac{f(x+h)-f(x)}{h} = \frac{1}{h}\left(\frac{2}{(x+h)^2} - \frac{2}{x^2}\right). Combine the two fractions over the common denominator x2(x+h)2x^2(x+h)^2, giving numerator 2x2−2(x+h)22x^2 - 2(x+h)^2 over x2(x+h)2hx^2(x+h)^2 h.

Expand the numerator: 2x2−2(x2+2xh+h2)=−4xh−2h2=−2h(2x+h)2x^2 - 2(x^2 + 2xh + h^2) = -4xh - 2h^2 = -2h(2x+h). Dividing by hh cancels one factor of hh, leaving −2(2x+h)x2(x+h)2.\frac{-2(2x+h)}{x^2(x+h)^2}. Taking the limit as h→0h \to 0 gives f′(x)=−4xx4=−4x3.f'(x) = \frac{-4x}{x^4} = -\frac{4}{x^3}.

Sign of the derivative expression

Hint
Step-by-step solution

The final derivative was found to be negative four divided by x cubed, so the constant multiplying the fraction is negative. This came from simplifying negative four x over x to the fourth power. Recognizing this sign helps confirm the derivative decreases the function's value as x increases for positive x. The answer is negative.

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