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

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The discussion focuses on finding the derivative dy/dx for the equation (2x - 10 + e)^(y + 8) = e^(x - 6) at the point (5, -9). The user initially miscalculated the derivative, leading to confusion about the correct answer, which is (2 + e)/e. The correct application of logarithmic differentiation and the treatment of e as a constant were clarified, resolving the misunderstanding regarding the differentiation process.

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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?
 
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Nope said:

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:

<br /> \frac{d}{d x}\left(2 x - 10 + e\right) \stackrel{?}{=} 2 + e<br />

EDIT:

The green term is fine.
 


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

ty
 

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