Just what the doctor ordered. Challenged me but not overwhelming. Good lesson for me: Show my work step by step -- my arithmetic errors sprout like dandelions.
Just last week I started taking Jason's Linear Algebra course for the second time. Why? Just for kicks and because he is a heck of a teacher. Wish I was a teenager again. Hey kids, opportunity is banging the door here. Answer the door you won't regret it. Five stars.
In say y = sin(x^2) We should just make x^2 =u. Then y = sin u Then dy/du = cos u and dy = cos u . du = cos (x^2) .du. Dividng both sides by dx gives du/dx = cos(x^2)*du/dx But du =dx^2. So du/dx =2x Final answer is dy/dx = cos(x^2).2x.
![The Chain Rule for Derivatives in Calculus - [1-5]](http://i.ytimg.com/vi/HpymsAXhJAM/mqdefault.jpg)
Just what the doctor ordered. Challenged me but not overwhelming. Good lesson for me: Show my work step by step -- my arithmetic errors sprout like dandelions.
Just last week I started taking Jason's Linear Algebra course for the second time. Why? Just for kicks and because he is a heck of a teacher. Wish I was a teenager again. Hey kids, opportunity is banging the door here. Answer the door you won't regret it. Five stars.
thanks, it really helped me, i wish my teacher could explain like this, they literally provide only basic then bombard us rock hard questions on exam
Greatest Math Teacher Ever!
Amazing sir, Love from India
Great review for me. Thank you!
Thanks alot
Nice lesson,
Nice 👍
In say y = sin(x^2)
We should just make x^2 =u.
Then y = sin u
Then dy/du = cos u and dy = cos u . du = cos (x^2) .du.
Dividng both sides by dx gives du/dx = cos(x^2)*du/dx
But du =dx^2. So du/dx =2x Final answer is dy/dx = cos(x^2).2x.