Is This the Correct Way to Differentiate y = x^(x^2 - 7)?

[Chadlee88]

i jst wanted to know if this is right. I need to find out the derivative of:
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)

 

[TD]

Your notation is a bit unclear to me, but the logarithm is a good idea:

$$y = x^{x^2 - 7} \Rightarrow y = \exp \left( {\ln \left( {x^{x^2 - 7} } \right)} \right) = \exp \left( {\left( {x^2 - 7} \right)\ln \left( x \right)} \right)$$

Can you find the derivative of that exponential?

 

[HallsofIvy]

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 $y= x^{x^2-7}$ so you rewrite it as
[tex]ln(y)= (x^2- 7)ln(x)[/itex]
The derivative of ln y with respect to y is $\frac{1}{y}$ but you want the derivative with respect to x- so use the chain rule:
$$\frac{d ln y}{dx}= \frac{d ln y}{dy}\frac{dy}{dx}= \frac{1}{y}\frac{dy}{dx}$$
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:
$$\frac{1}{y}y'= 2x ln x+ \frac{x^2- 7}{x}$$
and solve for y'.