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How do you find the derivative of # (7e^x)/(2e^x+1)#?

Answer 1

Paisley Johnson

#d/(d x)[(7e^x)/(2e^x+1)]=(7 e^x*l n(e))/(2e^x+1)-(14e^(2x)*l n(e))/(2e^2+1)^2#

#d/(d x)[(7e^x)/(2e^x+1)]=?#

#((7e^x)^'*(2e^x+1)-(2e^x+1)^' *7e^x)/((2e^x+1)^2)#

#=[7e^x*l n (e)*(2e^x+1)- ( 2*e^x * l n (e)*7e^x)]/(2e^x+1)^2#

#d/(d x)[(7e^x)/(2e^x+1)]=(7 e^x*l n(e))/(2e^x+1)-(7e^x*2e^x*l n(e))/(2e^2+1)^2#

#d/(d x)[(7e^x)/(2e^x+1)]=(7 e^x*l n(e))/(2e^x+1)-(14e^(2x)*l n(e))/(2e^x+1)^2#

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Answer 2

William Abdalla

To find the derivative of ( \frac{7e^x}{2e^x + 1} ), apply the quotient rule, which states that for functions ( u(x) ) and ( v(x) ):

[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} ]

Where ( u' ) represents the derivative of ( u(x) ) with respect to ( x ) and ( v' ) represents the derivative of ( v(x) ) with respect to ( x ). Apply the quotient rule to ( u(x) = 7e^x ) and ( v(x) = 2e^x + 1 ), and then simplify the expression.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.