The definition of derivative uses limits to find the exact instantaneous rate of change of a function. This page focuses on setting up the difference quotient, simplifying the expression, canceling the troublesome factor, and then evaluating the limit. These examples include polynomial, rational, radical, and point-slope derivative definition problems.
Notes
Practice Problems
\(\textbf{1)}\) \( f(x)=\frac{3}{x}, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{3}{x+h}-\frac{3}{x}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{3x-3(x+h)}{x(x+h)}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{-3h}{x(x+h)h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{-3}{x(x+h)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \frac{-3}{x^2}\)
The answer is \( f'(x)=\frac{-3}{x^2} \)
\(\textbf{2)}\) \( f(x)=3x^2+2x+1, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\left[3(x+h)^2+2(x+h)+1\right]-\left[3x^2+2x+1\right]}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{3x^2+6xh+3h^2+2x+2h+1-3x^2-2x-1}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{6xh+3h^2+2h}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0} 6x+3h+2\)
\(\,\,\,\,\,\,f'(x)=\displaystyle 6x+2\)
\(\text{The answer is } f'(x)=6x+2 \)
\(\textbf{3)}\) \( f(x)=x^2-3x+1, \) find \( f'(4) \) using the definition of derivative
\(\,\,\,\,\,\,f'(c)=\displaystyle \lim_{x\to c}\frac{f(x)-f(c)}{x-c}\)
\(\,\,\,\,\,\,f'(4)=\displaystyle \lim_{x\to4}\frac{f(x)-f(4)}{x-4}\)
\(\,\,\,\,\,\,f'(4)=\displaystyle \lim_{x\to4}\frac{\left(x^2-3x+1\right)-5}{x-4}\)
\(\,\,\,\,\,\,f'(4)=\displaystyle \lim_{x\to4}\frac{x^2-3x-4}{x-4}\)
\(\,\,\,\,\,\,f'(4)=\displaystyle \lim_{x\to4}\frac{(x+1)(x-4)}{x-4}\)
\(\,\,\,\,\,\,f'(4)=\displaystyle \lim_{x\to4}(x+1)\)
\(\,\,\,\,\,\,\)The answer is \( f'(4)=5 \)
\(\textbf{4)}\) \( f(x)=x^3-2x+1, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{(x+h)^3-2(x+h)+1-(x^3-2x+1)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{x^3+3x^2h+3xh^2+h^3-2x-2h+1-x^3+2x-1}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{3x^2h+3xh^2+h^3-2h}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}(3x^2+3xh+h^2-2)\)
\(\,\,\,\,\,\,f'(x)=\displaystyle 3x^2-2\)
The answer is \( f'(x)=3x^2-2 \)
\(\textbf{5)}\) Find the instantaneous slope of \( f(x)=x^2+x \) at \( x=3 \) using the definition of derivative at a point.
\(\,\,\,\,\,\,f'(3)=\displaystyle \lim_{x\to3}\frac{f(x)-f(3)}{x-3}\)
\(\,\,\,\,\,\,f'(3)=\displaystyle \lim_{x\to3}\frac{\left(x^2+x\right)-\left(3^2+3\right)}{x-3}\)
\(\,\,\,\,\,\,f'(3)=\displaystyle \lim_{x\to3}\frac{x^2+x-12}{x-3}\)
\(\,\,\,\,\,\,f'(3)=\displaystyle \lim_{x\to3}\frac{(x+4)(x-3)}{x-3}\)
\(\,\,\,\,\,\,f'(3)=\displaystyle \lim_{x\to3}(x+4)\)
\(\,\,\,\,\,\,f'(3)=7\)
The answer is \( f'(3)=7 \)
\(\textbf{6)}\) \( f(x)=x^2, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{(x+h)^2-x^2}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{x^2+2xh+h^2-x^2}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{h(2x+h)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}(2x+h)\)
\(\,\,\,\,\,\,f'(x)=\displaystyle 2x\)
The answer is \( f'(x)=2x \)
\(\textbf{7)}\) \( f(x)=\sqrt{x}, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{(x+h)-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \frac{1}{2\sqrt{x}}\)
The answer is \( f'(x)=\frac{1}{2\sqrt{x}} \)
\(\textbf{8)}\) \( f(x)=\frac{1}{x+2}, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{1}{x+h+2}-\frac{1}{x+2}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{x+2-(x+h+2)}{(x+h+2)(x+2)}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{-h}{(x+h+2)(x+2)}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{-1}{(x+h+2)(x+2)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \frac{-1}{(x+2)^2}\)
The answer is \( f'(x)=\frac{-1}{(x+2)^2} \)
\(\textbf{9)}\) Find the instantaneous slope of \( f(x)=\sqrt{x} \) at \( x=9 \) using the definition of derivative at a point.
\(\,\,\,\,\,\,f'(9)=\displaystyle \lim_{x\to9}\frac{f(x)-f(9)}{x-9}\)
\(\,\,\,\,\,\,f'(9)=\displaystyle \lim_{x\to9}\frac{\sqrt{x}-3}{x-9}\)
\(\,\,\,\,\,\,f'(9)=\displaystyle \lim_{x\to9}\frac{\sqrt{x}-3}{x-9}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+3}\)
\(\,\,\,\,\,\,f'(9)=\displaystyle \lim_{x\to9}\frac{x-9}{(x-9)(\sqrt{x}+3)}\)
\(\,\,\,\,\,\,f'(9)=\displaystyle \lim_{x\to9}\frac{1}{\sqrt{x}+3}\)
\(\,\,\,\,\,\,f'(9)=\displaystyle \frac{1}{6}\)
The answer is \( f'(9)=\frac{1}{6} \)
\(\textbf{10)}\) \( f(x)=4x-7, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\left(4(x+h)-7\right)-\left(4x-7\right)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{4x+4h-7-4x+7}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{4h}{h}\)
\(\,\,\,\,\,\,f'(x)=4\)
The answer is \( f'(x)=4 \)
\(\textbf{11)}\) \( f(x)=x^2+5x, \) find \( f'(2) \) using the definition of derivative at a point
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{f(x)-f(2)}{x-2}\)
\(\,\,\,\,\,\,f(2)=2^2+5(2)=14\)
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{x^2+5x-14}{x-2}\)
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{(x+7)(x-2)}{x-2}\)
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}(x+7)\)
\(\,\,\,\,\,\,f'(2)=9\)
The answer is \( f'(2)=9 \)
\(\textbf{12)}\) \( f(x)=x^2-9, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\left((x+h)^2-9\right)-\left(x^2-9\right)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{x^2+2xh+h^2-9-x^2+9}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{2xh+h^2}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}(2x+h)\)
\(\,\,\,\,\,\,f'(x)=2x\)
The answer is \( f'(x)=2x \)
\(\textbf{13)}\) \( f(x)=5x^2, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{5(x+h)^2-5x^2}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{5(x^2+2xh+h^2)-5x^2}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{10xh+5h^2}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}(10x+5h)\)
\(\,\,\,\,\,\,f'(x)=10x\)
The answer is \( f'(x)=10x \)
\(\textbf{14)}\) \( f(x)=\frac{2}{x}, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{2}{x+h}-\frac{2}{x}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{2x-2(x+h)}{x(x+h)}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{-2h}{x(x+h)h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{-2}{x(x+h)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle -\frac{2}{x^2}\)
The answer is \( f'(x)=-\frac{2}{x^2} \)
\(\textbf{15)}\) \( f(x)=x^3, \) find \( f'(1) \) using the definition of derivative at a point
\(\,\,\,\,\,\,f'(1)=\displaystyle \lim_{x\to1}\frac{f(x)-f(1)}{x-1}\)
\(\,\,\,\,\,\,f'(1)=\displaystyle \lim_{x\to1}\frac{x^3-1}{x-1}\)
\(\,\,\,\,\,\,f'(1)=\displaystyle \lim_{x\to1}\frac{(x-1)(x^2+x+1)}{x-1}\)
\(\,\,\,\,\,\,f'(1)=\displaystyle \lim_{x\to1}(x^2+x+1)\)
\(\,\,\,\,\,\,f'(1)=3\)
The answer is \( f'(1)=3 \)
\(\textbf{16)}\) \( f(x)=\sqrt{x+1}, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\cdot\frac{\sqrt{x+h+1}+\sqrt{x+1}}{\sqrt{x+h+1}+\sqrt{x+1}}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{(x+h+1)-(x+1)}{h\left(\sqrt{x+h+1}+\sqrt{x+1}\right)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{h}{h\left(\sqrt{x+h+1}+\sqrt{x+1}\right)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \frac{1}{2\sqrt{x+1}}\)
The answer is \( f'(x)=\frac{1}{2\sqrt{x+1}} \)
\(\textbf{17)}\) \( f(x)=\frac{1}{x-1}, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{1}{x+h-1}-\frac{1}{x-1}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{x-1-(x+h-1)}{(x+h-1)(x-1)}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\frac{-h}{(x+h-1)(x-1)}}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{-1}{(x+h-1)(x-1)}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle -\frac{1}{(x-1)^2}\)
The answer is \( f'(x)=-\frac{1}{(x-1)^2} \)
\(\textbf{18)}\) \( f(x)=2x^2-3x, \) find \( f'(x) \) using the definition of derivative
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{\left[2(x+h)^2-3(x+h)\right]-\left[2x^2-3x\right]}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{2x^2+4xh+2h^2-3x-3h-2x^2+3x}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}\frac{4xh+2h^2-3h}{h}\)
\(\,\,\,\,\,\,f'(x)=\displaystyle \lim_{h\to0}(4x+2h-3)\)
\(\,\,\,\,\,\,f'(x)=4x-3\)
The answer is \( f'(x)=4x-3 \)
\(\textbf{19)}\) Find the instantaneous slope of \( f(x)=\frac{1}{x} \) at \( x=2 \) using the definition of derivative at a point.
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{f(x)-f(2)}{x-2}\)
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{\frac{1}{x}-\frac{1}{2}}{x-2}\)
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{\frac{2-x}{2x}}{x-2}\)
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{-(x-2)}{2x(x-2)}\)
\(\,\,\,\,\,\,f'(2)=\displaystyle \lim_{x\to2}\frac{-1}{2x}\)
\(\,\,\,\,\,\,f'(2)=-\frac{1}{4}\)
The answer is \( f'(2)=-\frac{1}{4} \)
\(\textbf{20)}\) Find the instantaneous slope of \( f(x)=x^2-4x \) at \( x=5 \) using the definition of derivative at a point.
\(\,\,\,\,\,\,f'(5)=\displaystyle \lim_{x\to5}\frac{f(x)-f(5)}{x-5}\)
\(\,\,\,\,\,\,f(5)=5^2-4(5)=5\)
\(\,\,\,\,\,\,f'(5)=\displaystyle \lim_{x\to5}\frac{x^2-4x-5}{x-5}\)
\(\,\,\,\,\,\,f'(5)=\displaystyle \lim_{x\to5}\frac{(x-5)(x+1)}{x-5}\)
\(\,\,\,\,\,\,f'(5)=\displaystyle \lim_{x\to5}(x+1)\)
\(\,\,\,\,\,\,f'(5)=6\)
The answer is \( f'(5)=6 \)
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