How to find g'(x) when its given g(x)=6/x^2 + 1/3x + 2?

Question

Would it be g'(x)= -12/x + 1/3?

Answer 1

assuming I understood you correctly, g(x) = 6/(x^2) + 1/3(x) + 2

we calculate the derivative like so

d/dx g(x) = d/dx (6/(x^2)) + d/dx 1/3x + d/dx 2

the last two derivatives are equal to 1/3 and 0 respectively and d/dx 6/(x^2) = d/dx 6x^(-2) = 6(-2)x^-3 = -12x^(-3) = -12/(x^3) by the chain rule

and thus our derivative is g'(x) = -12/(x^3) + 1/3

I hope that helps

Answer 2

Assuming f(x) = 6/(x²)+(1/3)x

Write as 6x^-2+(1/3)x

Then f'(x) = -12x^-3 +1/3 (when you reduce -2 by 1 the new exponent is -3 (not -1).

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