Differentiating Natural Logarithms

[eutas1]

Please refer to the attachment - I am not sure how to get the answer that it is supposed to be...

Thank you!

 

[Deleted member 4993]

Please refer to the attachment - I am not sure how to get the answer that it is supposed to be...

Thank you!

Are you confused about transformation of the "last line of your work" to the "given answer"?

 

[skeeter]

problem statement says the function is

[MATH]y = \frac{\ln(x-5)}{2x+1} [/MATH]
you worked the derivative of

[MATH]y=\frac{\ln(x+5)}{2x+1}[/MATH]
,,,

[MATH]y’= \frac{(2x+1) \cdot \frac{1}{x-5} - 2 \cdot \ln(x-5)}{(2x+1)^2}[/MATH]
now, multiply numerator and denominator by [MATH](x-5)[/MATH] to clear the complex fraction in the numerator ... then split the resulting fraction

 

[eutas1]

problem statement says the function is

[MATH]y = \frac{\ln(x-5)}{2x+1} [/MATH]
you worked the derivative of

[MATH]y=\frac{\ln(x+5)}{2x+1}[/MATH]
,,,

[MATH]y’= \frac{(2x+1) \cdot \frac{1}{x-5} - 2 \cdot \ln(x-5)}{(2x+1)^2}[/MATH]
now, multiply numerator and denominator by [MATH](x-5)[/MATH] to clear the complex fraction in the numerator ... then split the resulting fraction

Ah, my mistake! I am stuck on the part where you split the resulting fraction... Please refer to my attachment

 

[skeeter]

[MATH]\frac{(2x+1) \cdot \frac{1}{x-5} - 2\ln(x-5)}{(2x+1)^2} \cdot \frac{x-5}{x-5}[/MATH]
[MATH]\frac{(2x+1) - 2(x-5)\ln(x-5)}{(2x+1)^2(x-5)}[/MATH]
[MATH]\frac{2x+1}{(2x+1)^2(x-5)} - \frac{2(x-5) \ln(x-5)}{(2x+1)^2(x-5)}[/MATH]
finish it ...

 

[eutas1]

[MATH]\frac{(2x+1) \cdot \frac{1}{x-5} - 2\ln(x-5)}{(2x+1)^2} \cdot \frac{x-5}{x-5}[/MATH]
[MATH]\frac{(2x+1) - 2(x-5)\ln(x-5)}{(2x+1)^2(x-5)}[/MATH]
[MATH]\frac{2x+1}{(2x+1)^2(x-5)} - \frac{2(x-5) \ln(x-5)}{(2x+1)^2(x-5)}[/MATH]
finish it ...

OHHHHH I see I see, thank you!!!