Question

Use definition of a derivative to compute the derivative of g(x)= 1/3x^2

Answer 1

If g(x) = 1/(3x²), the by the definition of the derivative, we have

`\displaystyle g'(x) = \lim_(h\to0) \frac{g(x+h) - g(x)}h`

`\displaystyle g'(x) = \lim_(h\to0) \frac{\frac1{3(x+h)^2} - \frac1{3x^2}}h`

`\displaystyle g'(x) = \lim_(h\to0) \frac{(x^2)/(3x^2(x+h)^2) - ((x+h)^2)/(3x^2(x+h)^2)}h`

`\displaystyle g'(x) = \lim_(h\to0) (x^2 - (x+h)^2)/(3x^2(x+h)^2h)`

`\displaystyle g'(x) = \lim_(h\to0) (x^2 - x^2 - 2xh - h^2)/(3x^2(x+h)^2h)`

`\displaystyle g'(x) = - \lim_(h\to0) (2xh + h^2)/(3x^2(x+h)^2h)`

`\displaystyle g'(x) = - \lim_(h\to0) (2x + h)/(3x^2(x+h)^2)`

`\displaystyle g'(x) = - (2x)/(3x^4) = \boxed{-\frac2{3x^3}}`