Find Answers
Get Study Tools
Get your homework help
23 Areas of Math We Help With
30k+ Questions answered
2 min avg time to find the answer
20b(4b3)3
I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example: 3 y 2 = 5 x 3 I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get: d d x ( 3 y 2 ) = d d x ( 5 x 3 ) 6 y d d x ( y ) = 15 x 2 d d x ( x ) 6 y d y d x = 15 x 2 d x d x 6 y d y d x = 15 x 2 d y d x = 15 x 2 6 y Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.
See answers (2)
I am having trouble finding a vector that is perpendicular to both a → = − 9 i ^ + 2 j ^ + 6 k ^ and b → = − 2 i ^ + j ^ + k ^ Would it just be 18 i + 2 j + 6 k ?
Prove that: tan(2tan−1(x))=2tan(tan−1(x)+tan−1(x3))My Attempt tan−1(x)=Ax=tan(A)Now, L.H.S =tan(2tan−1(x))=tan(2A)=2tan(A)1−tan2(A)=2x1−x2
See answers (1)
How do you find the intervals where the function f(x)=(2x−12)(x−3)2 is concave up and concave down?
Solve the simultaneous Linear equations using matrix inverse Method. 8x+3y=2 6x+2y=4
Let Z 1 and Z 2 be independent standard normal random variables and U 1 = Z 1 and U 2 = Z 1 + Z 2 . Are U 1 and U 2 independent? Why?
Solving inequality having logI am struggling to solve this inequality involving logarithm. How to find out values of n for which below inequality holds good: log 2 n n > 1 8
Attempting to factor (x2)+5x=0 via the grouping method.
Prices starting at $5/week., cancel anytime
Step-by-step solutions on your subject developed by experts