- #1

- 41

- 0

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

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)

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- Thread starter Chadlee88
- Start date

- #1

- 41

- 0

y = x^(x^2 - 7)

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

1/y =

x

y =

(x^2 + 2x - 7)

dy/dx =

(x^2 + 2x -7)

- #2

TD

Homework Helper

- 1,022

- 0

[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

I don't see how you got from x

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

[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

On the right side you want to differentiate (x

[tex]\frac{1}{y}y'= 2x ln x+ \frac{x^2- 7}{x}[/tex]

and solve for y'.

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