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f(V)=(V^(3))/(20)-7*V^(2)+265*V

f(V)=V320−7⋅V2+265⋅V f(V)=\frac{V^{3}}{20}-7 \cdot V^{2}+265 \cdot V

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Full solution

Q. f(V)=V320−7⋅V2+265⋅V f(V)=\frac{V^{3}}{20}-7 \cdot V^{2}+265 \cdot V
  1. Differentiate V3/20V^3/20: Differentiate the term V3/20V^3/20. Using the power rule, the derivative of V3V^3 is 3V23V^2. So, (V3/20)′=3V2/20(V^3/20)' = 3V^2/20.
  2. Differentiate −7V2-7V^2: Differentiate the term −7V2-7V^2. Using the power rule, the derivative of V2V^2 is 2V2V. So, (−7V2)′=−7×2V=−14V(-7V^2)' = -7 \times 2V = -14V.
  3. Differentiate 265V265V: Differentiate the term 265V265V.\newlineThe derivative of VV with respect to VV is 11.\newlineSo, (265V)′=265×1=265(265V)' = 265 \times 1 = 265.
  4. Combine derivatives: Combine the derivatives from steps 11 to 33.\newlinef′(V)=3V220−14V+265f'(V) = \frac{3V^2}{20} - 14V + 265.

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