Derivative of y = x^(x^2-7)

  • Thread starter Chadlee88
  • Start date
  • #1
41
0
i jst wanted to know if this is right. I need to find out the derivative of:
y = x^(x^2 - 7) :smile:

y = x^(x^2 - 7)

ln y = (x^2 -7) ln (x)

1/y = x^2 + 2x -7
x

y = x
(x^2 + 2x - 7)

dy/dx = -x^2 + 2x - 9 <----------- my answer
(x^2 + 2x -7)
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
Your notation is a bit unclear to me, but the logarithm is a good idea:

[tex]
y = x^{x^2 - 7} \Rightarrow y = \exp \left( {\ln \left( {x^{x^2 - 7} } \right)} \right) = \exp \left( {\left( {x^2 - 7} \right)\ln \left( x \right)} \right)
[/tex]

Can you find the derivative of that exponential?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
964
Beware of non-fixed fonts! Much better in Latex.
I don't see how you got from x2- 7 to x2+ 2x- 7 and I certainly don't see where that -x2+ 2x- 9 came from!
And, you seem to first solve for y, then it magically becomes y'.

You want to differentiate [itex]y= x^{x^2-7}[/itex] so you rewrite it as
[tex]ln(y)= (x^2- 7)ln(x)[/itex]
The derivative of ln y with respect to y is [itex]\frac{1}{y}[/itex] but you want the derivative with respect to x- so use the chain rule:
[tex]\frac{d ln y}{dx}= \frac{d ln y}{dy}\frac{dy}{dx}= \frac{1}{y}\frac{dy}{dx}[/tex]
but it is exactly dy/dx you want to find!
The left hand side is not just 1/y but is (1/y)y'.

On the right side you want to differentiate (x2- 7)ln x: use the product rule- (fg)'= f'g+ fg'. ((x2-7)ln(x))'= (x2-7)' ln(x)+ (x2-7)(ln x)'. The derivative of x2- 7 is 2x and the derivative of ln x is 1/x so ((x2-7)ln(x))'= (2x) ln(x)+ (x2-7)/x. Put those together:
[tex]\frac{1}{y}y'= 2x ln x+ \frac{x^2- 7}{x}[/tex]
and solve for y'.
 

Related Threads on Derivative of y = x^(x^2-7)

  • Last Post
Replies
4
Views
2K
Replies
5
Views
4K
  • Last Post
Replies
2
Views
820
  • Last Post
Replies
15
Views
23K
Replies
2
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
3
Views
2K
Replies
9
Views
2K
Replies
30
Views
8K
  • Last Post
Replies
5
Views
5K
Top