Are My Calculus Derivatives Correct?

  • Thread starter Thread starter eeyore81
  • Start date Start date
  • Tags Tags
    Calculus Grade 12
Click For Summary
SUMMARY

The forum discussion focuses on the calculation of derivatives for two functions: \( y = x^5 e^{-3x} \) and \( y = \frac{1 - e^{-x}}{1 + e^{2x}} \). The initial derivative calculations provided by the user were incorrect due to an extra "x" and missing terms. The corrected derivative for the first function is \( y' = (5x^4 e^{-3x}) + (x^5 - 3e^{-3x}) \). The second function's derivative also requires adjustments, including a missing minus sign and a squared term in the denominator.

PREREQUISITES
  • Understanding of calculus, specifically differentiation techniques.
  • Familiarity with exponential functions and their derivatives.
  • Knowledge of the product rule and quotient rule in calculus.
  • Ability to simplify algebraic expressions involving derivatives.
NEXT STEPS
  • Study the product rule for differentiation in calculus.
  • Learn the quotient rule for finding derivatives of rational functions.
  • Practice solving derivatives of exponential functions.
  • Review common mistakes in derivative calculations and how to avoid them.
USEFUL FOR

Students studying calculus, mathematics educators, and anyone seeking to improve their skills in differentiation and error-checking in derivative calculations.

eeyore81
Messages
2
Reaction score
0
For both questions i have to find dy/dx
1. y=x^5e^-3x

I got (5x^4 e^-3x) + (x^5 (-3xe^-3x-1))

some how that doesn't look right to me. not too sure...

2.
1-e^-x
y= --------
1+e^2x
My answer was:


(e^-x) - (1-e^-x) (2xe^2x-1)
y'= -----------------------------
1+e^2x


Thanks for your help! It's much appreciated!
 
Physics news on Phys.org
eeyore81 said:
For both questions i have to find dy/dx
1. y=x^5e^-3x
I got (5x^4 e^-3x) + (x^5 (-3xe^-3x-1))
some how that doesn't look right to me. not too sure...
2.
1-e^-x
y= --------
1+e^2x
My answer was: (e^-x) - (1-e^-x) (2xe^2x-1)
y'= -----------------------------
1+e^2x
Thanks for your help! It's much appreciated!

1.It's wrong.It is an "x" too much.Figure out where. :wink:
2.It's wrong.Again an "x" too much,a 'minus' missing,a whole term missing and a sqare on the denominator.Figure it out yourself.

Daniel.
 
First of all, thanks for replying, I'm still working on fixing the 2nd question, but i think i got the first one, the answer should be:


y' = (5x^4 e^-3x) + (x^5 -3e^-3x)
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
845
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
24
Views
4K