← Back to learning objective

Derivative of a sum involving a negative power and a monomial

Let p(x)=4x3−6x2p(x) = \dfrac{4}{x^3} - 6x^2. Find p′(x)p'(x).

Hint

First rewrite 4x3\dfrac{4}{x^3} as 4x−34x^{-3} so both terms are in the form c⋅xnc\cdot x^n. Then differentiate each term separately using the power rule and constant multiple rule, and combine the results using the difference rule.

Step-by-step solution

Rewrite p(x)=4x−3−6x2p(x) = 4x^{-3} - 6x^2 so that the power rule can be applied directly to each term. This avoids errors that commonly arise from differentiating a quotient involving xx in the denominator.

Differentiate term by term using the constant multiple and power rules: ddx[4x−3]=4(−3)x−4=−12x−4\dfrac{d}{dx}[4x^{-3}] = 4(-3)x^{-4} = -12x^{-4}, and ddx[6x2]=6(2)x1=12x\dfrac{d}{dx}[6x^2] = 6(2)x^{1} = 12x. Since p(x)p(x) is a difference of these two expressions, subtract the second derivative from the first.

Combining these results and rewriting the negative exponent as a fraction gives p′(x)=−12x−4−12x=−12x4−12xp'(x) = -12x^{-4} - 12x = -\dfrac{12}{x^4} - 12x.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap