Hello, paile!
By the way, the natural log is written , not .
Take logs: .
Differentiate implicitly: .
Then: .
Therefore: .
Do I still need to go any further?
No . . . Your work is excellent!
I came across this math problem,where i had to apply Implicit Differentiation to determine the derivative of y=x^e^-x^2.So in solving this problem,I introduced the natural log on either sides of the equation,to give Iny=Inx^e^-x^2,then appling one of the rules for logarithmic funtions,I made the exponent of my funtion to be the coefficient of Inx.Then i began differentating:
y=x^e^-x^2
Iny=Inx^e^-x^2
=e^-x^2Inx
1/y dy/dx=-2xe^-x^2Inx+e^-x^2/x .......................(A)
I understand that I came to (A) through the use of the Chain Rule,the I understant that I had to make dy/dx the subject of my formula,so as to have:
dy/dx=y(-2xe^-x^2Inx+e^-x^2/x)
Therefore this is the part where I experiance a problem,I substitute y by its initial value (i.e x^e^-x^2) to get:
dy/dx=x^e^-x^2(-2xe^-x^2Inx+e^-x^2/x) .........................(B).I arrived at (B) as my final solution,so I dont know if this is the correct solution then do I still need to go any further than I've come,do I need to proceed any further from my end point?