The normal line to a curve is the line that is perpendicular to the tangent line at a specific point. To find it, first find the derivative to get the tangent slope, then use the negative reciprocal to get the normal slope. These problems include polynomial, exponential, logarithmic, trigonometric, radical, and quotient-style functions.
Notes
Practice Questions
\(\small{\textbf{1)}}\) Find the equation of the normal line to the curve \(f(x) = x^3 – 4x + 1\) at the point \((2, 1)\). The equation of the normal line is \(y = -\frac{1}{8}x + \frac{5}{4}\)
\(\,\,\,\,\,f(x) = x^3 – 4x + 1\)
\(\,\,\,\,\,f'(x) = 3x^2 – 4\)
\(\,\,\,\,\,f'(2) = 3(2)^2 – 4 = 8\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(2)} = -\frac{1}{8}\)
\(\,\,\,\,\,f(2) = (2)^3 – 4(2) + 1 = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = -\frac{1}{8}(x – 2)\)
\(\,\,\,\,\,\text{Simplified: } y = -\frac{1}{8}x + \frac{5}{4}\)
\(\,\,\,\,\,f(x) = x^3 – 4x + 1\)
\(\,\,\,\,\,f'(x) = 3x^2 – 4\)
\(\,\,\,\,\,f'(2) = 3(2)^2 – 4 = 8\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(2)} = -\frac{1}{8}\)
\(\,\,\,\,\,f(2) = (2)^3 – 4(2) + 1 = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = -\frac{1}{8}(x – 2)\)
\(\,\,\,\,\,\text{Simplified: } y = -\frac{1}{8}x + \frac{5}{4}\)
\(\small{\textbf{2)}}\) Find the equation of the normal line to the curve \(f(x) = e^x\) at the point \((0, 1)\). The equation of the normal line is \(y = -x + 1\)
\(\,\,\,\,\,f(x) = e^x\)
\(\,\,\,\,\,f'(x) = e^x\)
\(\,\,\,\,\,f'(0) = e^0 = 1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(0)} = -1\)
\(\,\,\,\,\,f(0) = e^0 = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = -1(x – 0)\)
\(\,\,\,\,\,\text{Simplified: } y = -x + 1\)
\(\,\,\,\,\,f(x) = e^x\)
\(\,\,\,\,\,f'(x) = e^x\)
\(\,\,\,\,\,f'(0) = e^0 = 1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(0)} = -1\)
\(\,\,\,\,\,f(0) = e^0 = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = -1(x – 0)\)
\(\,\,\,\,\,\text{Simplified: } y = -x + 1\)
\(\small{\textbf{3)}}\) Find the equation of the normal line to the curve \(f(x) = \ln(x)\) at the point \((1, 0)\). The equation of the normal line is \(y = -x +1\)
\(\,\,\,\,\,f(x) = \ln(x)\)
\(\,\,\,\,\,f'(x) = \frac{1}{x}\)
\(\,\,\,\,\,f'(1) = \frac{1}{1} = 1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(1)} = -1\)
\(\,\,\,\,\,f(1) = \ln(1) = 0\)
\(\,\,\,\,\,\text{Point-slope form: } y – 0 = -1(x – 1)\)
\(\,\,\,\,\,\text{Simplified: } y = -x +1\)
\(\,\,\,\,\,f(x) = \ln(x)\)
\(\,\,\,\,\,f'(x) = \frac{1}{x}\)
\(\,\,\,\,\,f'(1) = \frac{1}{1} = 1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(1)} = -1\)
\(\,\,\,\,\,f(1) = \ln(1) = 0\)
\(\,\,\,\,\,\text{Point-slope form: } y – 0 = -1(x – 1)\)
\(\,\,\,\,\,\text{Simplified: } y = -x +1\)
\(\small{\textbf{4)}}\) Find the equation of the normal line to the curve \(f(x) = \sin(x)\) at the point \(\left(\frac{\pi}{2}, 1\right)\).
The equation of the normal line is \(x= \frac{\pi}{2}\)
\(\,\,\,\,\,f(x) = \sin(x)\)
\(\,\,\,\,\,f'(x) = \cos(x)\)
\(\,\,\,\,\,f’\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\)
\(\,\,\,\,\,\text{Slope of tangent line: } m = 0\)
\(\,\,\,\,\,\text{Slope of normal line: } m = \text{undefined (since } -\frac{1}{0} \text{ is undefined)}\)
\(\,\,\,\,\,\text{The normal line is vertical and passes through } x = \frac{\pi}{2}\)
\(\,\,\,\,\,\text{Equation: } x = \frac{\pi}{2}\)
\(\,\,\,\,\,f(x) = \sin(x)\)
\(\,\,\,\,\,f'(x) = \cos(x)\)
\(\,\,\,\,\,f’\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\)
\(\,\,\,\,\,\text{Slope of tangent line: } m = 0\)
\(\,\,\,\,\,\text{Slope of normal line: } m = \text{undefined (since } -\frac{1}{0} \text{ is undefined)}\)
\(\,\,\,\,\,\text{The normal line is vertical and passes through } x = \frac{\pi}{2}\)
\(\,\,\,\,\,\text{Equation: } x = \frac{\pi}{2}\)
\(\small{\textbf{5)}}\) Find the equation of the normal line to the curve \(f(x) = \frac{1}{x}\) at the point \((1, 1)\). The equation of the normal line is \(y = x \)
\(\,\,\,\,\,f(x) = \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = -\frac{1}{x^2}\)
\(\,\,\,\,\,f'(1) = -\frac{1}{1^2} = -1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(1)} = 1\)
\(\,\,\,\,\,f(1) = \frac{1}{1} = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = 1(x – 1)\)
\(\,\,\,\,\,\text{Simplified: } y = x \)
\(\,\,\,\,\,f(x) = \frac{1}{x}\)
\(\,\,\,\,\,f'(x) = -\frac{1}{x^2}\)
\(\,\,\,\,\,f'(1) = -\frac{1}{1^2} = -1\)
\(\,\,\,\,\,\text{Slope of normal line: } m = -\frac{1}{f'(1)} = 1\)
\(\,\,\,\,\,f(1) = \frac{1}{1} = 1\)
\(\,\,\,\,\,\text{Point-slope form: } y – 1 = 1(x – 1)\)
\(\,\,\,\,\,\text{Simplified: } y = x \)
\(\small{\textbf{6)}}\) Find the equation of the normal line to the curve \(f(x)=\sqrt{x}\) at the point \((4,2)\). The equation of the normal line is \(y=-4x+18\)
\(\,\,\,\,\,f(x)=\sqrt{x}\)
\(\,\,\,\,\,f(x)=x^{1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}x^{-1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\,\,f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(4)}=-4\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-4(x-4)\)
\(\,\,\,\,\,\text{Simplified: } y=-4x+18\)
\(\,\,\,\,\,f(x)=\sqrt{x}\)
\(\,\,\,\,\,f(x)=x^{1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}x^{-1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x}}\)
\(\,\,\,\,\,f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4}\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(4)}=-4\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-4(x-4)\)
\(\,\,\,\,\,\text{Simplified: } y=-4x+18\)
\(\small{\textbf{7)}}\) Find the equation of the normal line to the curve \(f(x)=x^2+\frac{1}{x}\) at the point \((1,2)\). The equation of the normal line is \(y=-x+3\)
\(\,\,\,\,\,f(x)=x^2+\frac{1}{x}\)
\(\,\,\,\,\,f(x)=x^2+x^{-1}\)
\(\,\,\,\,\,f'(x)=2x-x^{-2}\)
\(\,\,\,\,\,f'(x)=2x-\frac{1}{x^2}\)
\(\,\,\,\,\,f'(1)=2(1)-\frac{1}{1^2}=1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-1\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-1(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-x+3\)
\(\,\,\,\,\,f(x)=x^2+\frac{1}{x}\)
\(\,\,\,\,\,f(x)=x^2+x^{-1}\)
\(\,\,\,\,\,f'(x)=2x-x^{-2}\)
\(\,\,\,\,\,f'(x)=2x-\frac{1}{x^2}\)
\(\,\,\,\,\,f'(1)=2(1)-\frac{1}{1^2}=1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-1\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-1(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-x+3\)
\(\small{\textbf{8)}}\) Find the equation of the normal line to the curve \(f(x)=e^{2x}\) at the point \((0,1)\). The equation of the normal line is \(y=-\frac{1}{2}x+1\)
\(\,\,\,\,\,f(x)=e^{2x}\)
\(\,\,\,\,\,f'(x)=2e^{2x}\)
\(\,\,\,\,\,f'(0)=2e^{2(0)}=2\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-\frac{1}{2}\)
\(\,\,\,\,\,f(0)=e^{2(0)}=1\)
\(\,\,\,\,\,\text{Point-slope form: } y-1=-\frac{1}{2}(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{2}x+1\)
\(\,\,\,\,\,f(x)=e^{2x}\)
\(\,\,\,\,\,f'(x)=2e^{2x}\)
\(\,\,\,\,\,f'(0)=2e^{2(0)}=2\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-\frac{1}{2}\)
\(\,\,\,\,\,f(0)=e^{2(0)}=1\)
\(\,\,\,\,\,\text{Point-slope form: } y-1=-\frac{1}{2}(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{2}x+1\)
\(\small{\textbf{9)}}\) Find the equation of the normal line to the curve \(f(x)=\tan(x)\) at the point \((0,0)\). The equation of the normal line is \(y=-x\)
\(\,\,\,\,\,f(x)=\tan(x)\)
\(\,\,\,\,\,f'(x)=\sec^2(x)\)
\(\,\,\,\,\,f'(0)=\sec^2(0)=1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-1\)
\(\,\,\,\,\,f(0)=\tan(0)=0\)
\(\,\,\,\,\,\text{Point-slope form: } y-0=-1(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-x\)
\(\,\,\,\,\,f(x)=\tan(x)\)
\(\,\,\,\,\,f'(x)=\sec^2(x)\)
\(\,\,\,\,\,f'(0)=\sec^2(0)=1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-1\)
\(\,\,\,\,\,f(0)=\tan(0)=0\)
\(\,\,\,\,\,\text{Point-slope form: } y-0=-1(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-x\)
\(\small{\textbf{10)}}\) Find the equation of the normal line to the curve \(f(x)=\sqrt{x^2+3}\) at the point \((1,2)\). The equation of the normal line is \(y=-2x+4\)
\(\,\,\,\,\,f(x)=\sqrt{x^2+3}\)
\(\,\,\,\,\,f(x)=(x^2+3)^{1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}(x^2+3)^{-1/2}(2x)\)
\(\,\,\,\,\,f'(x)=\frac{x}{\sqrt{x^2+3}}\)
\(\,\,\,\,\,f'(1)=\frac{1}{\sqrt{1^2+3}}=\frac{1}{2}\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-2\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-2(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-2x+4\)
\(\,\,\,\,\,f(x)=\sqrt{x^2+3}\)
\(\,\,\,\,\,f(x)=(x^2+3)^{1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}(x^2+3)^{-1/2}(2x)\)
\(\,\,\,\,\,f'(x)=\frac{x}{\sqrt{x^2+3}}\)
\(\,\,\,\,\,f'(1)=\frac{1}{\sqrt{1^2+3}}=\frac{1}{2}\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-2\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-2(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-2x+4\)
\(\small{\textbf{11)}}\) Find the equation of the normal line to the curve \(f(x)=x^2+3x\) at the point \((1,4)\). The equation of the normal line is \(y=-\frac{1}{5}x+\frac{21}{5}\)
\(\,\,\,\,\,f(x)=x^2+3x\)
\(\,\,\,\,\,f'(x)=2x+3\)
\(\,\,\,\,\,f'(1)=2(1)+3=5\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-\frac{1}{5}\)
\(\,\,\,\,\,\text{Point-slope form: } y-4=-\frac{1}{5}(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{5}x+\frac{21}{5}\)
\(\,\,\,\,\,f(x)=x^2+3x\)
\(\,\,\,\,\,f'(x)=2x+3\)
\(\,\,\,\,\,f'(1)=2(1)+3=5\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-\frac{1}{5}\)
\(\,\,\,\,\,\text{Point-slope form: } y-4=-\frac{1}{5}(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{5}x+\frac{21}{5}\)
\(\small{\textbf{12)}}\) Find the equation of the normal line to the curve \(f(x)=x^4\) at the point \((1,1)\). The equation of the normal line is \(y=-\frac{1}{4}x+\frac{5}{4}\)
\(\,\,\,\,\,f(x)=x^4\)
\(\,\,\,\,\,f'(x)=4x^3\)
\(\,\,\,\,\,f'(1)=4(1)^3=4\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-\frac{1}{4}\)
\(\,\,\,\,\,\text{Point-slope form: } y-1=-\frac{1}{4}(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{4}x+\frac{5}{4}\)
\(\,\,\,\,\,f(x)=x^4\)
\(\,\,\,\,\,f'(x)=4x^3\)
\(\,\,\,\,\,f'(1)=4(1)^3=4\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-\frac{1}{4}\)
\(\,\,\,\,\,\text{Point-slope form: } y-1=-\frac{1}{4}(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{4}x+\frac{5}{4}\)
\(\small{\textbf{13)}}\) Find the equation of the normal line to the curve \(f(x)=\cos(x)\) at the point \((0,1)\). The equation of the normal line is \(x=0\)
\(\,\,\,\,\,f(x)=\cos(x)\)
\(\,\,\,\,\,f'(x)=-\sin(x)\)
\(\,\,\,\,\,f'(0)=-\sin(0)=0\)
\(\,\,\,\,\,\text{Slope of tangent line: } m=0\)
\(\,\,\,\,\,\text{Slope of normal line: } m=\text{undefined}\)
\(\,\,\,\,\,\text{The normal line is vertical and passes through }x=0\)
\(\,\,\,\,\,\text{Equation: } x=0\)
\(\,\,\,\,\,f(x)=\cos(x)\)
\(\,\,\,\,\,f'(x)=-\sin(x)\)
\(\,\,\,\,\,f'(0)=-\sin(0)=0\)
\(\,\,\,\,\,\text{Slope of tangent line: } m=0\)
\(\,\,\,\,\,\text{Slope of normal line: } m=\text{undefined}\)
\(\,\,\,\,\,\text{The normal line is vertical and passes through }x=0\)
\(\,\,\,\,\,\text{Equation: } x=0\)
\(\small{\textbf{14)}}\) Find the equation of the normal line to the curve \(f(x)=\ln(x+1)\) at the point \((0,0)\). The equation of the normal line is \(y=-x\)
\(\,\,\,\,\,f(x)=\ln(x+1)\)
\(\,\,\,\,\,f'(x)=\frac{1}{x+1}\)
\(\,\,\,\,\,f'(0)=\frac{1}{0+1}=1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-1\)
\(\,\,\,\,\,\text{Point-slope form: } y-0=-1(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-x\)
\(\,\,\,\,\,f(x)=\ln(x+1)\)
\(\,\,\,\,\,f'(x)=\frac{1}{x+1}\)
\(\,\,\,\,\,f'(0)=\frac{1}{0+1}=1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-1\)
\(\,\,\,\,\,\text{Point-slope form: } y-0=-1(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-x\)
\(\small{\textbf{15)}}\) Find the equation of the normal line to the curve \(f(x)=x^3+x\) at the point \((1,2)\). The equation of the normal line is \(y=-\frac{1}{4}x+\frac{9}{4}\)
\(\,\,\,\,\,f(x)=x^3+x\)
\(\,\,\,\,\,f'(x)=3x^2+1\)
\(\,\,\,\,\,f'(1)=3(1)^2+1=4\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-\frac{1}{4}\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-\frac{1}{4}(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{4}x+\frac{9}{4}\)
\(\,\,\,\,\,f(x)=x^3+x\)
\(\,\,\,\,\,f'(x)=3x^2+1\)
\(\,\,\,\,\,f'(1)=3(1)^2+1=4\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=-\frac{1}{4}\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=-\frac{1}{4}(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{4}x+\frac{9}{4}\)
\(\small{\textbf{16)}}\) Find the equation of the normal line to the curve \(f(x)=e^{-x}\) at the point \((0,1)\). The equation of the normal line is \(y=x+1\)
\(\,\,\,\,\,f(x)=e^{-x}\)
\(\,\,\,\,\,f'(x)=-e^{-x}\)
\(\,\,\,\,\,f'(0)=-e^0=-1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=1\)
\(\,\,\,\,\,\text{Point-slope form: } y-1=1(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=x+1\)
\(\,\,\,\,\,f(x)=e^{-x}\)
\(\,\,\,\,\,f'(x)=-e^{-x}\)
\(\,\,\,\,\,f'(0)=-e^0=-1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=1\)
\(\,\,\,\,\,\text{Point-slope form: } y-1=1(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=x+1\)
\(\small{\textbf{17)}}\) Find the equation of the normal line to the curve \(f(x)=\frac{x+1}{x}\) at the point \((1,2)\). The equation of the normal line is \(y=x+1\)
\(\,\,\,\,\,f(x)=\frac{x+1}{x}\)
\(\,\,\,\,\,f(x)=1+\frac{1}{x}\)
\(\,\,\,\,\,f'(x)=-\frac{1}{x^2}\)
\(\,\,\,\,\,f'(1)=-\frac{1}{1^2}=-1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=1\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=1(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=x+1\)
\(\,\,\,\,\,f(x)=\frac{x+1}{x}\)
\(\,\,\,\,\,f(x)=1+\frac{1}{x}\)
\(\,\,\,\,\,f'(x)=-\frac{1}{x^2}\)
\(\,\,\,\,\,f'(1)=-\frac{1}{1^2}=-1\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(1)}=1\)
\(\,\,\,\,\,\text{Point-slope form: } y-2=1(x-1)\)
\(\,\,\,\,\,\text{Simplified: } y=x+1\)
\(\small{\textbf{18)}}\) Find the equation of the normal line to the curve \(f(x)=\sqrt{x+5}\) at the point \((4,3)\). The equation of the normal line is \(y=-6x+27\)
\(\,\,\,\,\,f(x)=\sqrt{x+5}\)
\(\,\,\,\,\,f(x)=(x+5)^{1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}(x+5)^{-1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x+5}}\)
\(\,\,\,\,\,f'(4)=\frac{1}{2\sqrt{9}}=\frac{1}{6}\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(4)}=-6\)
\(\,\,\,\,\,\text{Point-slope form: } y-3=-6(x-4)\)
\(\,\,\,\,\,\text{Simplified: } y=-6x+27\)
\(\,\,\,\,\,f(x)=\sqrt{x+5}\)
\(\,\,\,\,\,f(x)=(x+5)^{1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}(x+5)^{-1/2}\)
\(\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x+5}}\)
\(\,\,\,\,\,f'(4)=\frac{1}{2\sqrt{9}}=\frac{1}{6}\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(4)}=-6\)
\(\,\,\,\,\,\text{Point-slope form: } y-3=-6(x-4)\)
\(\,\,\,\,\,\text{Simplified: } y=-6x+27\)
\(\small{\textbf{19)}}\) Find the equation of the normal line to the curve \(f(x)=\sin(2x)\) at the point \((0,0)\). The equation of the normal line is \(y=-\frac{1}{2}x\)
\(\,\,\,\,\,f(x)=\sin(2x)\)
\(\,\,\,\,\,f'(x)=2\cos(2x)\)
\(\,\,\,\,\,f'(0)=2\cos(0)=2\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-\frac{1}{2}\)
\(\,\,\,\,\,\text{Point-slope form: } y-0=-\frac{1}{2}(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{2}x\)
\(\,\,\,\,\,f(x)=\sin(2x)\)
\(\,\,\,\,\,f'(x)=2\cos(2x)\)
\(\,\,\,\,\,f'(0)=2\cos(0)=2\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(0)}=-\frac{1}{2}\)
\(\,\,\,\,\,\text{Point-slope form: } y-0=-\frac{1}{2}(x-0)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{2}x\)
\(\small{\textbf{20)}}\) Find the equation of the normal line to the curve \(f(x)=x^2-4x+6\) at the point \((3,3)\). The equation of the normal line is \(y=-\frac{1}{2}x+\frac{9}{2}\)
\(\,\,\,\,\,f(x)=x^2-4x+6\)
\(\,\,\,\,\,f'(x)=2x-4\)
\(\,\,\,\,\,f'(3)=2(3)-4=2\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(3)}=-\frac{1}{2}\)
\(\,\,\,\,\,\text{Point-slope form: } y-3=-\frac{1}{2}(x-3)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{2}x+\frac{9}{2}\)
\(\,\,\,\,\,f(x)=x^2-4x+6\)
\(\,\,\,\,\,f'(x)=2x-4\)
\(\,\,\,\,\,f'(3)=2(3)-4=2\)
\(\,\,\,\,\,\text{Slope of normal line: } m=-\frac{1}{f'(3)}=-\frac{1}{2}\)
\(\,\,\,\,\,\text{Point-slope form: } y-3=-\frac{1}{2}(x-3)\)
\(\,\,\,\,\,\text{Simplified: } y=-\frac{1}{2}x+\frac{9}{2}\)
See Related Pages\(\)
\(\bullet\text{Derivative Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)
\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)
\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)
\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)
\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)
\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)
\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)
\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)
\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)
\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)
\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)
\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)
\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)
\(\bullet\text{ Implicit Differentiation}\)
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\(\bullet\text{ Horizontal Tangent Line}\)
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\(\bullet\text{ Mean Value Theorem}\)
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\(\bullet\text{ Related Rates}\)
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\(\bullet\text{ Increasing and Decreasing Intervals}\)
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\(\bullet\text{ Intervals of concave up and down}\)
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\(\bullet\text{ Inflection Points}\)
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\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
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\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)