Question

Differentiate:
d/dx(1/x^3)

Answer 1

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Answer 2

Solution :

Quotient rule of differentiation :

$\boxed{\sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dx} - (u)\dfrac{d(v)}{dx}}{(v)^{2}}}}$

According to the given information, we get that the value of (u = 1) and (v = x³).

$\textsf{By using the qoutient rule and} \\ \textsf{substituting the values in it, we get :} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dx} - (u)\dfrac{d(v)}{dx}}{(v)^{2}}} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{(x^{3})\dfrac{d(1)}{dx} - (1)\dfrac{d(x^{3})}{dx}}{(x^{3})^{2}}} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{(x^{3})0 - \dfrac{d(x^{3})}{dx}}{(x^{3})^{2}}} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{- \dfrac{d(x^{3})}{dx}}{(x^{3})^{2}}} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{- [3 \times x^{(3 - 1)}]}{(x^{3})^{2}}} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{- 3x^{2}}{(x)^{6}}} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{- 3}{(x)^{(6 - 2)}}} \\ \\ \\ :\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{- 3}{(x)^{4}}} \\ \\ \\ \boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{1}{x^{3}}\bigg) = \dfrac{- 3}{(x)^{4}}}}$

Hence the derivative of the function 1/x³ is -3/x⁴.

Knowledge required :

• Derivative of a constant term is 0.
• Power rule of differentiation : d(x^n)/dx = nx^(n - 1).