Differentiating (2x - 10 + e)^(y + 8à = e^(x - 6)

[Nope]

Homework Statement

(2x-10+e)^(y+8)=e^(x-6)
value of dy/dx at the point, (5,-9)

Homework Equations

The Attempt at a Solution

(y+8)ln(2x-10+e)=x-6

y' ln(2x-10+e)+(y+8)(2+e)/(2x-10+e)=1

ln(2*5-10+e)=1

y'+(-9+8)(2+e)/e=1
y'-(2+e)/e=1
y'=1+(2+e)/e

but the answer is (2+e)/e, I don't know where i went wrong..
derivative of ln(e^(x-6)), should be 1, right?

 

[Dickfore]

Homework Statement

(2x-10+e)^(y+8)=e^(x-6)
value of dy/dx at the point, (5,-9)

Homework Equations

The Attempt at a Solution

(y+8)ln(2x-10+e)=x-6

y' ln(2x-10+e)+(y+8)(2+e)/(2x-10+e)=1

ln(2*5-10+e)=1

y'+(-9+8)(2+e)/e=1
y'-(2+e)/e=1
y'=1+(2+e)/e

but the answer is (2+e)/e, I don't know where i went wrong..
derivative of ln(e^(x-6)), should be 1, right?

For the red term. Since when:

$$\frac{d}{d x}\left(2 x - 10 + e\right) \stackrel{?}{=} 2 + e$$

EDIT:

The green term is fine.

 

[Nope]

ok , i see it now, e is a constant, that's why..

ty