Find the Derivative of the function y = ln [(e^θ) / (10 + e^θ)]?

Question

Find the Derivative of the function y = ln [(e^θ) / (10 + e^θ)]

The answer given is 10 / (10 + e^θ)



I think this is how to do it?

= ln (e^θ) - ln (10 + e^θ)

??????? I'm not sure


Thanks

Answer 1

y = ln [(e^θ) / (10 + e^θ)]



differentiating with respect to θ, we have



dy/dθ = [(10 + e^θ)/(e^θ)]*[(e^θ)*(e^θ) - (10 + e^θ)*(e^θ)]/ (10 + e^θ)^2...................use the chain rule(first the derivative of the logarithm and then use quotient rule)



dy/dθ = [e^2θ - 10 - e^2θ]/[(e^θ)* (10 + e^θ)]



dy/dθ = -10 / [(e^θ)* (10 + e^θ)]

Answer 2

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Answer 3

for me.. the answer is 1

Answer 4

Note, it is easier for me to type x for theta so we start with:



y = In[e^(x) / (10 + e^(x)) ]



We use the rule that if y = In[f(x)] then dy/dx = f'(x)/f(x). Hence we have



f(x) = e^(x) / (10 + e^x) and we need to find f'(x)



Let y =- e^(x) / (10 + e^x) and let u = e^x and v = (10 + e^x). Then:



du/dx = e(x) and dv/dx = e(x)



By the Quotient rule: dy/dx = [v*du/dx - u*dv/dx]/v^2. Hence



dy/dx = (10 + e^x) * e^(x) - e^(x)*e^x)]/ (10 + e^x)^2



dy/dx = 10e^(x) / (10 + e^(x))^2



So having found f'(x) we can now go back to finding the derivative of y = In[e^(x)/ (10 + e^(x)) using the rule given at the start.



dy/dx = f'(x)/f(x)



we have f'(x) = 10e^(x)/(10 + e^(x))^2 and f (x) =[e^(x)/(10 + e^(x)]



Hence dy/dx = 10e^(x)/(10 + e^(x))^2 * (10 + e(x)) / e^(x)



Cancelling gives:



dy/dx = 10/ (10 + e^(x))



which is the answer you give when you replace theta for x

Answer 5

Not sure,

Maybe try: chain rule?



ln x = 1/x (substitute)



e^θ = e^θ (e^kx = ke^kx constant is 1)



10 + e^θ= e^θ (constants drop)



So chain rule says f(g(x)) -> dy/dx f(g(x)) * dydx g(x) -----or something like that



[ 1/(e^θ) . e^θ ] - [ 1/(e^θ) . e^θ ]



does that equal 0?



Hope that helps



EDIT:

OOPs! I left out the 10 + from the substitution!

[ 1/(e^θ) . e^θ ] - [ 1/(10 + e^θ) . e^θ ]

so it's 1 - [ 1/(10 + e^θ) . e^θ ]

Answer 6

y = ln (e^θ) - ln (10 + e^θ)

ln (e^θ) = θ

The equation becomes:



y = θ - ln(10+ e^θ)

dy/dθ = 1 - [1/(10+ e^θ)] d/dθ [ e^θ]



dy/dθ = 1 - e^θ/(10+ e^θ)

Answer 7

You could definitely start this way; ln (e^θ) is just θ, so



y = θ - ln (10 + e^θ), and dy/dθ = 1 - (10 + e^θ)^-1 * (e^θ) (Chain rule)



dy/dθ = 1 - e^θ/(10 + e^θ) = (10 + e^θ)/(10 + e^θ) - e^θ/(10 + e^θ) = 10/(10 + e^θ).