Finding a tangent line

joshmeyer

New member
Joined
Mar 20, 2022
Messages
19
1648063298213.png
I have absolutely no idea what I've done wrong, and unlike my last post, it's not that I didn't read instructions.
 
View attachment 31799
I have absolutely no idea what I've done wrong, and unlike my last post, it's not that I didn't read instructions.
It seems like you're trying to use the definition of derivative to find f'(x). You forgot that you'd have to take the limit as h->0, and you didn't apply the definition correctly. Are you required to use the definition of limit? It can be easily done by the power rule.
 
Last edited:
It seems like you're trying to use the definition of derivative to find f'(x). You forgot that you'd have to take the limit as h->0, and you didn't apply the definition correctly. Are you required to use the definition of limit? It can be easily done by the power rule.
I sadly have not been taught the power rule yet, but as it's an online course the only thing I have to show is the answer. If I could learn said power rule to skip what's 5 steps that would be great

It's time consuming to expand binomials past the power of 3
 
I sadly have not been taught the power rule yet, but as it's an online course the only thing I have to show is the answer. If I could learn said power rule to skip what's 5 steps that would be great

It's time consuming to expand binomials past the power of 3
I'm a bit surprised that you haven't learned the power rule, which is one of the most basic rules of taking derivatives, yet you're doing tangent lines. Anyhow, since you haven't learned the power rule yet, we can stick with what you learned
[math]f'(x)=\lim_{h\to 0}\frac{-2(x+h)^3-(-2x^3)}{h}[/math]Use the binomial theorem Subhotosh Khan provided in post#4 to continue.
 
ooooooooooohhhhhhh I see what I did wrong. I get it now. Looking forward to learning the power rule. I've always been the type of person who can just accept math rules, and trying to explain why the rule works has always confused me.
 
Top