Notes

Questions
\(\textbf{1)}\) Find the equation of the tangent line to the curve \( f(x)=x^3+3x^2-x \) at the point \( (2,18) \)
The answer is \( y=23x-28 \)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^3+3x^2-x\)
\(\,\,\,\,\,\,f'(x)=3x^2+6x-1\)
\(\,\,\,\,\,\,f'(2)=3(2)^2+6(2)-1\)
\(\,\,\,\,\,\,f'(2)=12+12-1\)
\(\,\,\,\,\,\,f'(2)=23\)
\(\,\,\,\,\,\,\text{Slope at } (2,18) \text{ is }m=23\)
\(\,\,\,\text{Step 2: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\,\text{Point Slope Formula: }y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\,y-18=23(x-2)\)
\(\,\,\,\,\,\,y-18=23x-46\)
\(\,\,\,\,\,\,\text{The answer is } y=23x-28 \)
\(\textbf{2)}\) Find the equation of the tangent line to the curve \( f(x)=x\sqrt{x} \) at the point \((4,8) \)
The answer is \( y=3x-4 \)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,\,\,\, f(x)=x\sqrt{x} \)
\(\,\,\,\,\,\,\,\,\, f(x)=x^1 \cdot x^{1/2} \)
\(\,\,\,\,\,\,\,\,\, f(x)=x^{1+1/2} \)
\(\,\,\,\,\,\,\,\,\, f(x)=x^{3/2} \)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{3}{2}x^{\left(3/2 – 1\right)} \)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{3}{2}x^{1/2} \)
\(\,\,\,\,\,\,\,\,\, f'(4)=\frac{3}{2}(4)^{1/2} \)
\(\,\,\,\,\,\,\,\,\, f'(4)=\frac{3}{2}(2) \)
\(\,\,\,\,\,\,\,\,\, f'(4)=3 \)
\(\,\,\,\,\,\,\,\,\,\text{Slope at } (4,8) \text{ is }m=3\)
\(\,\,\,\text{Step 2: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\,\,\,\,\text{Point Slope Formula: }y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\,\,\,\,y-8=3(x-4)\)
\(\,\,\,\,\,\,\,\,\,y-8=3x-12\)
\(\,\,\,\,\,\,\,\,\,y=3x-4\)
\(\,\,\,\)The answer is \( y=3x-4 \)
\(\textbf{3)}\) Find the equation of the tangent line to the curve \( f(x)=x^2-6x+4 \) and parallel to the line \(y=-4x+9 \)
The answer is \( y=-4x+3 \)
\(\,\,\,\text{Step 1: Parallel lines have the same slope.}\)
\(\,\,\,\,\,\,\,\,\, \text{The slope of the line } y=-4x+9 \text{ is } m=-4\)
\(\,\,\,\text{Step 2: Take the derivative of } f(x)=x^2-6x+4\)
\(\,\,\,\,\,\,\,\,\, f'(x)=2x-6 \)
\(\,\,\,\,\,\,\,\,\, f'(x)=-4 \)
\(\,\,\,\,\,\,\,\,\, 2x-6=-4 \)
\(\,\,\,\,\,\,\,\,\, x=1 \)
\(\,\,\,\text{Step 3: Find the y-coordinate at } x=1 \)
\(\,\,\,\,\,\,\,\,\, f(1)=1^2-6(1)+4=-1\)
\(\,\,\,\text{Step 4: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\,\,\,\, y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\,\,\,\, y-(-1)=-4(x-1)\)
\(\,\,\,\,\,\,\,\,\, y+1=-4x+4\)
\(\,\,\,\,\,\,\,\,\, y=-4x+3\)
\(\,\,\,\,\,\,\,\,\,\text{The answer is } y=-4x+3\)
\(\textbf{4)}\) Find the equation of the tangent line to the curve \( f(x)=\frac{e^x}{x} \) at the point \( (1,e) \)
The answer is \( y=e \)
\(\,\,\,\text{Step 1: Find the derivative using the quotient rule.}\)
\(\,\,\,\,\,\,\,\,\, f(x)=\frac{e^x}{x} \)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{x(e^x)-(e^x)(1)}{x^2}\)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{xe^x-e^x}{x^2}\)
\(\,\,\,\,\,\,\,\,\, f'(1)=\frac{1(e^1)-e^1}{1^2}\)
\(\,\,\,\,\,\,\,\,\, f'(1)=\frac{e-e}{1}=0 \)
\(\,\,\,\text{Step 2: Use the slope } m=0 \text{ and the point } (1,e) \)
\(\,\,\,\,\,\,\,\,\, y-e=0(x-1)\)
\(\,\,\,\,\,\,\,\,\, y=e \)
\(\,\,\,\,\,\,\,\,\,\text{The answer is } y=e \)
\(\textbf{5)}\) Find the equation of the tangent line to the curve \( f(x)=5e^x \cos{x} \) at the point \( (0,5) \)
The answer is \( y=5x+5 \)
\(\,\,\,\text{Step 1: Use the product rule for the derivative.}\)
\(\,\,\,\,\,\, f(x)=5e^x \cos{x} \)
\(\,\,\,\,\,\, f'(x)=5e^x \cos{x}-5e^x \sin{x} \)
\(\,\,\,\,\,\, f'(0)=5e^0 \cos{0}-5e^0 \sin{0} \)
\(\,\,\,\,\,\, f'(0)=5(1)(1)-5(1)(0)=5 \)
\(\,\,\,\text{Step 2: Use the slope } m=5 \text{ and the point } (0,5) \)
\(\,\,\,\,\,\, y-5=5(x-0)\)
\(\,\,\,\,\,\, y-5=5x\)
\(\,\,\,\,\,\, y=5x+5 \)
\(\,\,\,\,\,\,\text{The answer is } y=5x+5 \)
\(\textbf{6)}\) Find the equation of the tangent line to the curve \( f(x)=5\sqrt{x}(x+4) \) at the point \( (4,80) \)
The answer is \( y=20x\)
\(\,\,\,\text{Step 1: Distribute } 5x^{1/2} \text{ across } (x+4).\)
\(\,\,\,\,\,\, f(x)=5x^{1/2}(x)+5x^{1/2}(4)\)
\(\,\,\,\,\,\, f(x)=5x^{3/2}+20x^{1/2}\)
\(\,\,\,\text{Step 2: Differentiate using the power rule.}\)
\(\,\,\,\,\,\, f'(x)=\frac{15}{2}x^{1/2}+10x^{-1/2}\)
\(\,\,\,\text{Step 3: Evaluate } f'(x) \text{ at } x=4.\)
\(\,\,\,\,\,\, f'(4)=\frac{15}{2}(4)^{1/2}+10(4)^{-1/2}\)
\(\,\,\,\,\,\, f'(4)=\frac{15}{2}(2)+10(1/2)\)
\(\,\,\,\,\,\, f'(4)=15+5\)
\(\,\,\,\,\,\, f'(4)=20\)
\(\,\,\,\,\,\, \text{Slope at } (4,80) \text{ is } m=20\)
\(\,\,\,\text{Step 4: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\, y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\, y-80=20(x-4)\)
\(\,\,\,\,\,\, y-80=20x-80\)
\(\,\,\,\,\,\, y=20x\)
\(\,\,\,\,\,\, \text{The answer is } y=20x\)
See Related Pages\(\)
In Summary
Tangent lines are a key concept in calculus. The slope of a tangent line is same as the instantaneous slope (or derivative) of the graph at that point. We can find the equation of the tangent line by using point slope formula \(y-y_0=m\left(x-x_0\right)\), where we use the derivative value for the slope and the point of tangency as the point \(\left(x_0,y_0\right)\).