# Derivative of exponential function

1. Dec 6, 2012

### domyy

1. The problem statement, all variables and given/known data

Use implicit differentiation to find dy/dx.

2. Relevant equations

xey - 10x + 3y = 0

3. The attempt at a solution

= [xey + ey(y)'] - (10x)' + (3y)' = 0

= xey + ey(y') - 10 + 3(y') = 0

= y' (ey + 3) = 10 - xey

= y' = 10 - xey/ ey + 3y

However, my book says the answer: 10 - ey/ xey + 3y.

What am I doing wrong?

Thanks!

2. Dec 7, 2012

### Staff: Mentor

The first thing wrong is starting the line below with =. By putting that = at the start of each line, you are more likely to lose track of the two sides of the equation you started with.
The more serious mistake is above. You didn't take the derivative of xey correctly. d/dx(xey) = ey + xey*y'. Do you see why?
No, I believe their answer is (10 - ey)/(xey + 3). Note the parentheses, and the 3 instead of the 3y that you had.

3. Dec 7, 2012

### domyy

About taking the derivative wrongly, my question is: How does that differ from y= xe^x - e^x?

Because I was trying to follow the same thinking when solving the problem in question.

This is how I solved it ( and I got it right):

= xe^x + e^x (x)' - e^x(x)'
= xe^x + e^x - e^x
= e^x ( x -1 + 1)
= xe^x

Thanks!

Last edited: Dec 7, 2012
4. Dec 7, 2012

### Staff: Mentor

Don't start your work with =.
In this problem,
d/dx(xex) = x *d/dx(ex) + d/dx(x) * ex = xex + 1*ex = ex(x + 1)

The previous problem is different because you need to use the chain rule.
d/dx(xey) = x * d/dx(ey) + d/dx(x) * ey
= x * d/dy(ey) * dy/dx + d/dx(x) * ey = xey*y' + 1*ey = ey(xy' + 1)

5. Dec 7, 2012

### domyy

Ok, I am going to try to solve some other problems to practice it.

Last edited: Dec 7, 2012