The power rule is one of the most important derivative rules in calculus. It says that when differentiating a power of \(x\), you bring the exponent down in front and subtract 1 from the exponent. These problems include whole-number exponents, negative exponents, fractional exponents, radicals, constants, and expressions that need to be rewritten before using the rule.
Notes
Questions
Find the derivative
\(\textbf{1)}\) \(f(x)=\displaystyle\frac{1}{x^4}\) The answer is \(-\displaystyle\frac{4}{x^5}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{1}{x^4}\)
\(\,\,\,\,\,f(x)=x^{-4}\)
\(\,\,\,\,\,f'(x)=-4x^{-5}\)
\(\,\,\,\,\,f'(x)=-\displaystyle\frac{4}{x^5}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{4}{x^5}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{1}{x^4}\)
\(\,\,\,\,\,f(x)=x^{-4}\)
\(\,\,\,\,\,f'(x)=-4x^{-5}\)
\(\,\,\,\,\,f'(x)=-\displaystyle\frac{4}{x^5}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{4}{x^5}\)
\(\textbf{2)}\) \(f(x)=x^5\) The answer is \(5x^4\)
\(\,\,\,\,\,f(x)=x^5\)
\(\,\,\,\,\,f'(x)=5x^{5-1}\)
\(\,\,\,\,\,f'(x)=5x^4\)
\(\,\,\,\,\,\)The answer is \(5x^4\)
\(\,\,\,\,\,f(x)=x^5\)
\(\,\,\,\,\,f'(x)=5x^{5-1}\)
\(\,\,\,\,\,f'(x)=5x^4\)
\(\,\,\,\,\,\)The answer is \(5x^4\)
\(\textbf{3)}\) \(f(x)=x\) The answer is \(1\)
\(\,\,\,\,\,f(x)=x\)
\(\,\,\,\,\,f'(x)=1x^{1-1}\)
\(\,\,\,\,\,f'(x)=1x^0\)
\(\,\,\,\,\,f'(x)=1\)
\(\,\,\,\,\,\)The answer is \(1\)
\(\,\,\,\,\,f(x)=x\)
\(\,\,\,\,\,f'(x)=1x^{1-1}\)
\(\,\,\,\,\,f'(x)=1x^0\)
\(\,\,\,\,\,f'(x)=1\)
\(\,\,\,\,\,\)The answer is \(1\)
\(\textbf{4)}\) \(f(x)=\displaystyle\frac{3}{x}\) The answer is \(-\displaystyle\frac{3}{x^2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{3}{x}\)
\(\,\,\,\,\,f(x)=3x^{-1}\)
\(\,\,\,\,\,f'(x)=-1\cdot3x^{-1-1}\)
\(\,\,\,\,\,f'(x)=-\displaystyle\frac{3}{x^2}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{3}{x^2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{3}{x}\)
\(\,\,\,\,\,f(x)=3x^{-1}\)
\(\,\,\,\,\,f'(x)=-1\cdot3x^{-1-1}\)
\(\,\,\,\,\,f'(x)=-\displaystyle\frac{3}{x^2}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{3}{x^2}\)
\(\textbf{5)}\) \(f(x)=\displaystyle\frac{1}{x}\) The answer is \(-\displaystyle\frac{1}{x^2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{1}{x}\)
\(\,\,\,\,\,f(x)=x^{-1}\)
\(\,\,\,\,\,f'(x)=-1x^{-1-1}\)
\(\,\,\,\,\,f'(x)=-\displaystyle\frac{1}{x^2}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{1}{x^2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{1}{x}\)
\(\,\,\,\,\,f(x)=x^{-1}\)
\(\,\,\,\,\,f'(x)=-1x^{-1-1}\)
\(\,\,\,\,\,f'(x)=-\displaystyle\frac{1}{x^2}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{1}{x^2}\)
\(\textbf{6)}\) \(f(x)=3x^4-2x+8\) The answer is \(12x^3-2\)
\(\,\,\,\,\,f(x)=3x^4-2x+8\)
\(\,\,\,\,\,f'(x)=4\cdot3x^{4-1}-2\)
\(\,\,\,\,\,f'(x)=12x^3-2\)
\(\,\,\,\,\,\)The answer is \(12x^3-2\)
\(\,\,\,\,\,f(x)=3x^4-2x+8\)
\(\,\,\,\,\,f'(x)=4\cdot3x^{4-1}-2\)
\(\,\,\,\,\,f'(x)=12x^3-2\)
\(\,\,\,\,\,\)The answer is \(12x^3-2\)
\(\textbf{7)}\) \(f(x)=\displaystyle\frac{1}{6}x^3\) The derivative is \(f'(x)=\displaystyle\frac{x^2}{2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{1}{6}x^3\)
\(\,\,\,\,\,f'(x)=3\cdot\displaystyle\frac{1}{6}x^{3-1}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{x^2}{2}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{x^2}{2}\)
\(\,\,\,\,\,f(x)=\displaystyle\frac{1}{6}x^3\)
\(\,\,\,\,\,f'(x)=3\cdot\displaystyle\frac{1}{6}x^{3-1}\)
\(\,\,\,\,\,f'(x)=\displaystyle\frac{x^2}{2}\)
\(\,\,\,\,\,\)The derivative is \(f'(x)=\displaystyle\frac{x^2}{2}\)
\(\textbf{8)}\) \(f(x)=2^{30}\) The answer is \(f'(x)=0\)
\(\,\,\,\,\,f(x)=2^{30}\)
\(\,\,\,\,\,\text{Since }2^{30}\text{ is a constant, its derivative is }0.\)
\(\,\,\,\,\,f'(x)=0\)
\(\,\,\,\,\,\)The answer is \(f'(x)=0\)
\(\,\,\,\,\,f(x)=2^{30}\)
\(\,\,\,\,\,\text{Since }2^{30}\text{ is a constant, its derivative is }0.\)
\(\,\,\,\,\,f'(x)=0\)
\(\,\,\,\,\,\)The answer is \(f'(x)=0\)
\(\textbf{9)}\) \(f(x)=x^{30}\) The answer is \(f'(x)=30x^{29}\)
\(\,\,\,\,\,f(x)=x^{30}\)
\(\,\,\,\,\,f'(x)=30x^{30-1}\)
\(\,\,\,\,\,f'(x)=30x^{29}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=30x^{29}\)
\(\,\,\,\,\,f(x)=x^{30}\)
\(\,\,\,\,\,f'(x)=30x^{30-1}\)
\(\,\,\,\,\,f'(x)=30x^{29}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=30x^{29}\)
\(\textbf{10)}\) \(f(x)=5\sqrt{x^5}+\frac{10}{x}-x^{300}\) The answer is \(f'(x)=\frac{25}{2}x^{3/2}-10x^{-2}-300x^{299}\)
\(\,\,\,\,\,f(x)=5\sqrt{x^5}+\frac{10}{x}-x^{300}\)
\(\,\,\,\,\,f(x)=5x^{5/2}+10x^{-1}-x^{300}\)
\(\,\,\,\,\,f'(x)=\frac{5}{2} \cdot 5x^{5/2-1}+(-1)10x^{-1-1}-300x^{300-1}\)
\(\,\,\,\,\,f'(x)=\frac{25}{2}x^{3/2}-10x^{-2}-300x^{299}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\frac{25}{2}x^{3/2}-10x^{-2}-300x^{299}\)
\(\,\,\,\,\,f(x)=5\sqrt{x^5}+\frac{10}{x}-x^{300}\)
\(\,\,\,\,\,f(x)=5x^{5/2}+10x^{-1}-x^{300}\)
\(\,\,\,\,\,f'(x)=\frac{5}{2} \cdot 5x^{5/2-1}+(-1)10x^{-1-1}-300x^{300-1}\)
\(\,\,\,\,\,f'(x)=\frac{25}{2}x^{3/2}-10x^{-2}-300x^{299}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=\frac{25}{2}x^{3/2}-10x^{-2}-300x^{299}\)
\(\textbf{11)}\) \(f(x)=\sqrt{x}\) The answer is \(\displaystyle f'(x)=\frac{1}{2}x^{-1/2}\)
\(\,\,\,\,\,f(x)=\sqrt{x}=x^{1/2}\)
\(\,\,\,\,\,f'(x)=\tfrac{1}{2}x^{1/2-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{1}{2}x^{-1/2}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle f'(x)=\frac{1}{2}x^{-1/2}\)
\(\,\,\,\,\,f(x)=\sqrt{x}=x^{1/2}\)
\(\,\,\,\,\,f'(x)=\tfrac{1}{2}x^{1/2-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{1}{2}x^{-1/2}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle f'(x)=\frac{1}{2}x^{-1/2}\)
\(\textbf{12)}\) \(f(x)=\displaystyle\frac{7}{x^4}\) The answer is \(-\displaystyle\frac{28}{x^5}\)
\(\,\,\,\,\,f(x)=\frac{7}{x^4}=7x^{-4}\)
\(\,\,\,\,\,f'(x)=-4\cdot7x^{-4-1}\)
\(\,\,\,\,\,f'(x)=-\frac{28}{x^5}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{28}{x^5}\)
\(\,\,\,\,\,f(x)=\frac{7}{x^4}=7x^{-4}\)
\(\,\,\,\,\,f'(x)=-4\cdot7x^{-4-1}\)
\(\,\,\,\,\,f'(x)=-\frac{28}{x^5}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{28}{x^5}\)
\(\textbf{13)}\) \(f(x)=x^{2/3}\) The answer is \(\displaystyle \frac{2}{3}x^{-1/3}\)
\(\,\,\,\,\,f(x)=x^{2/3}\)
\(\,\,\,\,\,f'(x)=\tfrac{2}{3}x^{2/3-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{2}{3}x^{-1/3}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{2}{3}x^{-1/3}\)
\(\,\,\,\,\,f(x)=x^{2/3}\)
\(\,\,\,\,\,f'(x)=\tfrac{2}{3}x^{2/3-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{2}{3}x^{-1/3}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{2}{3}x^{-1/3}\)
\(\textbf{14)}\) \(f(x)=4x^{-5}-3x^{1/2}+9\) The answer is \(-20x^{-6}-\tfrac{3}{2}x^{-1/2}\)
\(\,\,\,\,\,f(x)=4x^{-5}-3x^{1/2}+9\)
\(\,\,\,\,\,f'(x)=-5\cdot4x^{-6}-\tfrac{1}{2}\cdot3x^{-1/2}+0\)
\(\,\,\,\,\,f'(x)=-20x^{-6}-\tfrac{3}{2}x^{-1/2}\)
\(\,\,\,\,\,\)The answer is \(-20x^{-6}-\tfrac{3}{2}x^{-1/2}\)
\(\,\,\,\,\,f(x)=4x^{-5}-3x^{1/2}+9\)
\(\,\,\,\,\,f'(x)=-5\cdot4x^{-6}-\tfrac{1}{2}\cdot3x^{-1/2}+0\)
\(\,\,\,\,\,f'(x)=-20x^{-6}-\tfrac{3}{2}x^{-1/2}\)
\(\,\,\,\,\,\)The answer is \(-20x^{-6}-\tfrac{3}{2}x^{-1/2}\)
\(\textbf{15)}\) \(f(x)=\displaystyle\frac{5}{2}x^{7/3}\) The answer is \(\displaystyle \frac{35}{6}x^{4/3}\)
\(\,\,\,\,\,f(x)=\tfrac{5}{2}x^{7/3}\)
\(\,\,\,\,\,f'(x)=\tfrac{7}{3}\cdot\tfrac{5}{2}x^{7/3-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{35}{6}x^{4/3}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{35}{6}x^{4/3}\)
\(\,\,\,\,\,f(x)=\tfrac{5}{2}x^{7/3}\)
\(\,\,\,\,\,f'(x)=\tfrac{7}{3}\cdot\tfrac{5}{2}x^{7/3-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{35}{6}x^{4/3}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{35}{6}x^{4/3}\)
\(\textbf{16)}\) \(f(x)=3\sqrt[4]{x^{9}}\) The answer is \(\displaystyle \frac{27}{4}x^{5/4}\)
\(\,\,\,\,\,f(x)=3x^{9/4}\)
\(\,\,\,\,\,f'(x)=3\cdot\tfrac{9}{4}x^{9/4-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{27}{4}x^{5/4}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{27}{4}x^{5/4}\)
\(\,\,\,\,\,f(x)=3x^{9/4}\)
\(\,\,\,\,\,f'(x)=3\cdot\tfrac{9}{4}x^{9/4-1}\)
\(\,\,\,\,\,f'(x)=\tfrac{27}{4}x^{5/4}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{27}{4}x^{5/4}\)
\(\textbf{17)}\) \(f(x)=x^{-1/3}-\displaystyle\frac{2}{\sqrt{x}}\) The answer is \(-\displaystyle\frac{1}{3}x^{-4/3}+x^{-3/2}\)
\(\,\,\,\,\,f(x)=x^{-1/3}-2x^{-1/2}\)
\(\,\,\,\,\,f'(x)=-\tfrac{1}{3}x^{-4/3}-2\cdot\left(-\tfrac{1}{2}\right)x^{-3/2}\)
\(\,\,\,\,\,f'(x)=-\tfrac{1}{3}x^{-4/3}+x^{-3/2}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{1}{3}x^{-4/3}+x^{-3/2}\)
\(\,\,\,\,\,f(x)=x^{-1/3}-2x^{-1/2}\)
\(\,\,\,\,\,f'(x)=-\tfrac{1}{3}x^{-4/3}-2\cdot\left(-\tfrac{1}{2}\right)x^{-3/2}\)
\(\,\,\,\,\,f'(x)=-\tfrac{1}{3}x^{-4/3}+x^{-3/2}\)
\(\,\,\,\,\,\)The answer is \(-\displaystyle\frac{1}{3}x^{-4/3}+x^{-3/2}\)
\(\textbf{18)}\) \(f(x)=\displaystyle\frac{1}{5}x^5+\frac{1}{3}x^3+\frac{1}{2}x^{-2}\) The answer is \(x^4+x^2-x^{-3}\)
\(\,\,\,\,\,f(x)=\tfrac{1}{5}x^5+\tfrac{1}{3}x^3+\tfrac{1}{2}x^{-2}\)
\(\,\,\,\,\,f'(x)=5\cdot\tfrac{1}{5}x^{4}+3\cdot\tfrac{1}{3}x^{2}+(-2)\cdot\tfrac{1}{2}x^{-3}\)
\(\,\,\,\,\,f'(x)=x^4+x^2-x^{-3}\)
\(\,\,\,\,\,\)The answer is \(x^4+x^2-x^{-3}\)
\(\,\,\,\,\,f(x)=\tfrac{1}{5}x^5+\tfrac{1}{3}x^3+\tfrac{1}{2}x^{-2}\)
\(\,\,\,\,\,f'(x)=5\cdot\tfrac{1}{5}x^{4}+3\cdot\tfrac{1}{3}x^{2}+(-2)\cdot\tfrac{1}{2}x^{-3}\)
\(\,\,\,\,\,f'(x)=x^4+x^2-x^{-3}\)
\(\,\,\,\,\,\)The answer is \(x^4+x^2-x^{-3}\)
\(\textbf{19)}\) \(f(x)=\displaystyle\frac{6}{\sqrt[3]{x^2}}\) The answer is \(-4x^{-5/3}\)
\(\,\,\,\,\,f(x)=6x^{-2/3}\)
\(\,\,\,\,\,f'(x)=-\tfrac{2}{3}\cdot6x^{-5/3}\)
\(\,\,\,\,\,f'(x)=-4x^{-5/3}\)
\(\,\,\,\,\,\)The answer is \(-4x^{-5/3}\)
\(\,\,\,\,\,f(x)=6x^{-2/3}\)
\(\,\,\,\,\,f'(x)=-\tfrac{2}{3}\cdot6x^{-5/3}\)
\(\,\,\,\,\,f'(x)=-4x^{-5/3}\)
\(\,\,\,\,\,\)The answer is \(-4x^{-5/3}\)
\(\textbf{20)}\) \(f(x)=x^{12}+\displaystyle\frac{1}{x^3}+\sqrt[4]{x}-\displaystyle\frac{1}{\sqrt{x^7}}+9\) The answer is \(f'(x)=12x^{11}-3x^{-4}+\tfrac{1}{4}x^{-3/4}+\tfrac{7}{2}x^{-9/2}\)
\(\,\,\,\,\,f(x)=x^{12}+x^{-3}+x^{1/4}-x^{-7/2}+9\)
\(\,\,\,\,\,f'(x)=12x^{11}-3x^{-4}+\tfrac{1}{4}x^{-3/4}+\tfrac{7}{2}x^{-9/2}+0\)
\(\,\,\,\,\,f'(x)=12x^{11}-3x^{-4}+\tfrac{1}{4}x^{-3/4}+\tfrac{7}{2}x^{-9/2}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=12x^{11}-3x^{-4}+\tfrac{1}{4}x^{-3/4}+\tfrac{7}{2}x^{-9/2}\)
\(\,\,\,\,\,f(x)=x^{12}+x^{-3}+x^{1/4}-x^{-7/2}+9\)
\(\,\,\,\,\,f'(x)=12x^{11}-3x^{-4}+\tfrac{1}{4}x^{-3/4}+\tfrac{7}{2}x^{-9/2}+0\)
\(\,\,\,\,\,f'(x)=12x^{11}-3x^{-4}+\tfrac{1}{4}x^{-3/4}+\tfrac{7}{2}x^{-9/2}\)
\(\,\,\,\,\,\)The answer is \(f'(x)=12x^{11}-3x^{-4}+\tfrac{1}{4}x^{-3/4}+\tfrac{7}{2}x^{-9/2}\)
See Related Pages\(\)
\(\bullet\text{ Derivative Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)
\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)
\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)
\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)
\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)
\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)
\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)
\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)
\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)
\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)
\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)
\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)
\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)
\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)
\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)