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help please

Guest Jun 1, 2017
 #1
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\(f(x)=\frac{3cotx}{2+e^x}\\ f(x)=\frac{3}{tanx(2+e^x)}\\ f(x)=3[(tanx)(2+e^x)]^{-1}\\ f'(x)=-3[(tanx)(2+e^x)]^{-2}\times [sec^2x(2+e^x)+e^xtanx]\\ f'(x)=\dfrac{-3[sec^2x(2+e^x)+e^xtanx]}{[(tanx)(2+e^x)]^{2}}\\ f'(x)=\dfrac{-3sec^2x(2+e^x)-3e^xtanx}{(tan^2x)(2+e^x)^{2}}\\\)

 

Any questions?

Melody  Jun 1, 2017
 #2
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+1

Possible derivation:
d/dx((3 cot(x))/(2 + e^x))


Factor out constants:
 = 3 (d/dx((cot(x))/(2 + e^x)))


Use the quotient rule, d/dx(u/v) = (v ( du)/( dx) - u ( dv)/( dx))/v^2, where u = cot(x) and v = e^x + 2:
 = 3 ((2 + e^x) d/dx(cot(x)) - cot(x) d/dx(2 + e^x))/(2 + e^x)^2


Differentiate the sum term by term:
 = (3 ((2 + e^x) (d/dx(cot(x))) - d/dx(2) + d/dx(e^x) cot(x)))/(2 + e^x)^2


The derivative of 2 is zero:
 = (3 ((2 + e^x) (d/dx(cot(x))) - cot(x) (d/dx(e^x) + 0)))/(2 + e^x)^2


Simplify the expression:
 = (3 (-(cot(x) (d/dx(e^x))) + (2 + e^x) (d/dx(cot(x)))))/(2 + e^x)^2


The derivative of e^x is e^x:
 = (3 ((2 + e^x) (d/dx(cot(x))) - e^x cot(x)))/(2 + e^x)^2


The derivative of cot(x) is -csc^2(x):
Answer: | = (3 (-(e^x cot(x)) + (2 + e^x) -csc(x)^2))/(2 + e^x)^2

Guest Jun 1, 2017

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