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Given: f(x) = 2x3 / x4. Determine the derivative of f(x) by using first principles. Show all working. 2.2. Use differentiation rules to determine the derivatives of the following, leaving your answer with positive exponents and in surd form where applicable.

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Answer

−2x2\mathbf{-\frac{2}{x^2}}

Graph of f(x) = (2x³)/x⁴

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Step-by-step solution

6 steps

to solve Question 2.1, we will use the definition of the derivative from first principles.

  1. Simplify the given function f(x)f(x).

    f(x)=2x3x4=2x3−4=2x−1=2xf(x) = \frac{2x^3}{x^4} = 2x^{3-4} = 2x^{-1} = \frac{2}{x}

  2. Write down the formula for the derivative from first principles.

    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

  3. Substitute f(x)f(x) and f(x+h)f(x+h) into the formula.

    Since f(x)=2xf(x) = \frac{2}{x}, then f(x+h)=2x+hf(x+h) = \frac{2}{x+h}.

    f′(x)=lim⁡h→02x+h−2xhf'(x) = \lim_{h \to 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}

  4. Simplify the numerator by finding a common denominator.

    2x+h−2x=2x−2(x+h)x(x+h)\frac{2}{x+h} - \frac{2}{x} = \frac{2x - 2(x+h)}{x(x+h)} =2x−2x−2hx(x+h)= \frac{2x - 2x - 2h}{x(x+h)} =−2hx(x+h)= \frac{-2h}{x(x+h)}

  5. Substitute the simplified numerator back into the limit expression.

    f′(x)=lim⁡h→0−2hx(x+h)hf'(x) = \lim_{h \to 0} \frac{\frac{-2h}{x(x+h)}}{h} f′(x)=lim⁡h→0−2hh⋅x(x+h)f'(x) = \lim_{h \to 0} \frac{-2h}{h \cdot x(x+h)}

  6. Cancel hh and evaluate the limit.

    f′(x)=lim⁡h→0−2x(x+h)f'(x) = \lim_{h \to 0} \frac{-2}{x(x+h)}

    Now, substitute h=0h=0:

    f′(x)=−2x(x+0)f'(x) = \frac{-2}{x(x+0)} f′(x)=−2x2f'(x) = \frac{-2}{x^2}

    The derivative of f(x)f(x) using first principles is −2x2\boxed{\mathbf{-\frac{2}{x^2}}}.

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