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Derivative of a sum of power functions

Let g(x)=1x+3x2−5x.g(x) = \frac{1}{x} + \frac{3}{x^2} - 5x. Rewrite each term as a power of xx, then use the constant multiple and sum/difference rules together with the power rule to find g′(x)g'(x).

Hint

First rewrite 1x\frac{1}{x} as x−1x^{-1} and 3x2\frac{3}{x^2} as 3x−23x^{-2}. Then differentiate term by term, applying the power rule to each term and keeping constant multiples fixed.

Step-by-step solution

Rewrite g(x)g(x) using negative exponents: g(x)=x−1+3x−2−5x.g(x) = x^{-1} + 3x^{-2} - 5x. This form makes each term ready for the power rule, since dividing by a power of xx is the same as multiplying by a negative power of xx.

Differentiate each term separately using the sum/difference rule, applying the power rule ddx[xn]=nxn−1\frac{d}{dx}[x^n] = nx^{n-1} to each piece and the constant multiple rule where needed: ddx[x−1]=−x−2,ddx[3x−2]=−6x−3,ddx[−5x]=−5.\frac{d}{dx}[x^{-1}] = -x^{-2}, \quad \frac{d}{dx}[3x^{-2}] = -6x^{-3}, \quad \frac{d}{dx}[-5x] = -5. Adding these results gives g′(x)=−x−2−6x−3−5g'(x) = -x^{-2} - 6x^{-3} - 5, which can be rewritten as g′(x)=−1x2−6x3−5.g'(x) = -\frac{1}{x^2} - \frac{6}{x^3} - 5.

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