Andymath.com features free videos, notes, and practice problems with answers! Printable pages make math easy. Are you ready to be a mathmagician?

Questions

\(\textbf{1)}\) Find the equation of the tangent line to the curve \( f(x)=x^3+3x^2-x \) at the point \( (2,18) \)

Show Answer

The answer is \( y=23x-28 \)

Show Work

\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)

\(\,\,\,\,\,\,f(x)=x^3+3x^2-x\)

\(\,\,\,\,\,\,f'(x)=3x^2+6x-1\)

\(\,\,\,\,\,\,f'(2)=3(2)^2+6(2)-1\)

\(\,\,\,\,\,\,f'(2)=12+12-1\)

\(\,\,\,\,\,\,f'(2)=23\)

\(\,\,\,\,\,\,\text{Slope at } (2,18) \text{ is }m=23\)

\(\,\,\,\text{Step 2: Use the point and slope to find the equation of the line.}\)

\(\,\,\,\,\,\,\text{Point Slope Formula: }y-y_1=m(x-x_1)\)

\(\,\,\,\,\,\,y-18=23(x-2)\)

\(\,\,\,\,\,\,y-18=23x-46\)

\(\,\,\,\,\,\,\text{The answer is } y=23x-28 \)

\(\textbf{2)}\) Find the equation of the tangent line to the curve \( f(x)=x\sqrt{x} \) at the point \((4,8) \)

Show Answer

The answer is \( y=3x-4 \)

Show Work

\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)

\(\,\,\,\,\,\,\,\,\, f(x)=x\sqrt{x} \)

\(\,\,\,\,\,\,\,\,\, f(x)=x^{3/2} \)

\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{3}{2}x^{1/2} \)

\(\,\,\,\,\,\,\,\,\, f'(4)=\frac{3}{2}(4)^{1/2} \)

\(\,\,\,\,\,\,\,\,\, f'(4)=\frac{3}{2}(2) \)

\(\,\,\,\,\,\,\,\,\, f'(4)=3 \)

\(\,\,\,\,\,\,\,\,\,\text{Slope at } (4,8) \text{ is }m=3\)

\(\,\,\,\text{Step 2: Use the point and slope to find the equation of the line.}\)

\(\,\,\,\,\,\,\,\,\,\text{Point Slope Formula: }y-y_1=m(x-x_1)\)

\(\,\,\,\,\,\,\,\,\,y-8=3(x-4)\)

\(\,\,\,\,\,\,\,\,\,y-8=3x-12\)

\(\,\,\,\,\,\,\,\,\,y=3x-4\)

\(\,\,\,\)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 \(y=-4x+9 \)

Show Answer

The answer is \( y=-4x+3 \)

Show Work

\(\,\,\,\text{Step 1: Parallel lines have the same slope.}\)

\(\,\,\,\,\,\,\,\,\, \text{The slope of the line } y=-4x+9 \text{ is } m=-4\)

\(\,\,\,\text{Step 2: Take the derivative of } f(x)=x^2-6x+4\)

\(\,\,\,\,\,\,\,\,\, f'(x)=2x-6 \)

\(\,\,\,\,\,\,\,\,\, f'(x)=-4 \)

\(\,\,\,\,\,\,\,\,\, 2x-6=-4 \)

\(\,\,\,\,\,\,\,\,\, x=1 \)

\(\,\,\,\text{Step 3: Find the y-coordinate at } x=1 \)

\(\,\,\,\,\,\,\,\,\, f(1)=1^2-6(1)+4=-1\)

\(\,\,\,\text{Step 4: Use the point and slope to find the equation of the line.}\)

\(\,\,\,\,\,\,\,\,\, y-y_1=m(x-x_1)\)

\(\,\,\,\,\,\,\,\,\, y-(-1)=-4(x-1)\)

\(\,\,\,\,\,\,\,\,\, y+1=-4x+4\)

\(\,\,\,\,\,\,\,\,\, y=-4x+3\)

\(\,\,\,\,\,\,\,\,\,\text{The answer is } y=-4x+3\)

\(\textbf{4)}\) Find the equation of the tangent line to the curve \( f(x)=\frac{e^x}{x} \) at the point \( (1,e) \)

Show Answer

The answer is \( y=e \)

Show Work

\(\,\,\,\text{Step 1: Find the derivative using the quotient rule.}\)

\(\,\,\,\,\,\,\,\,\, f(x)=\frac{e^x}{x} \)

\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{x(e^x)-(e^x)(1)}{x^2}\)

\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{xe^x-e^x}{x^2}\)

\(\,\,\,\,\,\,\,\,\, f'(1)=\frac{1(e^1)-e^1}{1^2}\)

\(\,\,\,\,\,\,\,\,\, f'(1)=\frac{e-e}{1}=0 \)

\(\,\,\,\text{Step 2: Use the slope } m=0 \text{ and the point } (1,e) \)

\(\,\,\,\,\,\,\,\,\, y-e=0(x-1)\)

\(\,\,\,\,\,\,\,\,\, y=e \)

\(\,\,\,\,\,\,\,\,\,\text{The answer is } y=e \)

\(\textbf{5)}\) Find the equation of the tangent line to the curve \( f(x)=5e^x \cos⁡{x} \) at the point \( (0,5) \)

Show Answer

The answer is \( y=5x+5 \)

Show Work

\(\,\,\,\text{Step 1: Use the product rule for the derivative.}\)

\(\,\,\,\,\,\, f(x)=5e^x \cos{x} \)

\(\,\,\,\,\,\, f'(x)=5e^x \cos{x}-5e^x \sin{x} \)

\(\,\,\,\,\,\, f'(0)=5e^0 \cos{0}-5e^0 \sin{0} \)

\(\,\,\,\,\,\, f'(0)=5(1)(1)-5(1)(0)=5 \)

\(\,\,\,\text{Step 2: Use the slope } m=5 \text{ and the point } (0,5) \)

\(\,\,\,\,\,\, y-5=5(x-0)\)

\(\,\,\,\,\,\, y-5=5x\)

\(\,\,\,\,\,\, y=5x+5 \)

\(\,\,\,\,\,\,\text{The answer is } y=5x+5 \)

\(\textbf{6)}\) Find the equation of the tangent line to the curve \( f(x)=5\sqrt{x}(x+4) \) at the point \( (4,80) \)

Show Answer

The answer is \( y=20x\)

Show Work

\(\,\,\,\text{Step 1: Distribute } 5x^{1/2} \text{ across } (x+4).\)

\(\,\,\,\,\,\, f(x)=5x^{1/2}(x)+5x^{1/2}(4)\)

\(\,\,\,\,\,\, f(x)=5x^{3/2}+20x^{1/2}\)

\(\,\,\,\text{Step 2: Differentiate using the power rule.}\)

\(\,\,\,\,\,\, f'(x)=\frac{15}{2}x^{1/2}+10x^{-1/2}\)

\(\,\,\,\text{Step 3: Evaluate } f'(x) \text{ at } x=4.\)

\(\,\,\,\,\,\, f'(4)=\frac{15}{2}(4)^{1/2}+10(4)^{-1/2}\)

\(\,\,\,\,\,\, f'(4)=\frac{15}{2}(2)+10(1/2)\)

\(\,\,\,\,\,\, f'(4)=15+5\)

\(\,\,\,\,\,\, f'(4)=20\)

\(\,\,\,\,\,\, \text{Slope at } (4,80) \text{ is } m=20\)

\(\,\,\,\text{Step 4: Use the point and slope to find the equation of the line.}\)

\(\,\,\,\,\,\, y-y_1=m(x-x_1)\)

\(\,\,\,\,\,\, y-80=20(x-4)\)

\(\,\,\,\,\,\, y-80=20x-80\)

\(\,\,\,\,\,\, y=20x\)

\(\,\,\,\,\,\, \text{The answer is } y=20x\)

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)

\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)

\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)

\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)

\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)

\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)

\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)

\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)

\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)

\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)

\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)

\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)

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 \(\left(x_0,y_0\right)\).

About Andymath.com

Andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. If you have any requests for additional content, please contact Andy at tutoring@andymath.com. He will promptly add the content.

Topics cover Elementary Math, Middle School, Algebra, Geometry, Algebra 2/Pre-calculus/Trig, Calculus and Probability/Statistics. In the future, I hope to add Physics and Linear Algebra content.

Visit me on Youtube, Tiktok, Instagram and Facebook. Andymath content has a unique approach to presenting mathematics. The clear explanations, strong visuals mixed with dry humor regularly get millions of views. We are open to collaborations of all types, please contact Andy at tutoring@andymath.com for all enquiries. To offer financial support, visit my Patreon page. Let’s help students understand the math way of thinking!

Thank you for visiting. How exciting!