What is the second derivative of #x/(2+ e^x)#?

Answer 1
We can use the quotient rule, here, which states that, for #y=f(x)/g(x)#,
#(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#
Thus, as #f(x)=x# and #g(x)=2+e^x#,
#(dy)/(dx)=(1*(2+e^x)-x(e^x))/(2+e^x)^2=color(green)((2+e^x-xe^x)/(2+e^x)^2)#
Now, we can rewrite this first derivative as a product: #(2+e^x-xe^x)(2+e^x)^-2#
Applying the product rule, which states that when #y=f(x)g(x)#,
#(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#
It'll be easier for us to find #f'(x)# and #g'(x)# before proceeding to the whole derivation, so we do not get lost.
#f'(x)=0+e^x+color(red)(1*e^x+x*e^x)#, thus #color(green)(f'(x)=2e^x+xe^x)# Note that the red part was the demanded chain rule for #xe^x#.
Now, #g'(x)# demands chain rule, which states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#.
Renaming #u=2+e^x#, we have #g(u)=u^-2#

So,

#(dy)/(du)=-2u^-3# and #(du)/(dx)=e^x#

Aggregating them in a multiplication as stated by chain rule:

#(dy)/(dx)=(-2u^-3)(e^x)=(-2(2+e^x)^-3)(e^x)#
#g'(x)=color(green)((-2(e^x))/(2+e^x)^3)#
Now that we have #f(x)#, #f'(x)#, #g(x)# and #g'(x)#, we can go to product rule:
#(d^2y)/(dx^2)=(e^x+e^x+xe^x)(e^x+2)^-2+(2+e^x+xe^x)(-2e^x(e^x+2)^-3)#
#(d^2y)/(dx^2)=(2e^x+xe^x)/(e^x+2)^2+((2+e^x+xe^x)(-2e^x))/(e^x+2)^3#
To operate this sum, we have that the lowest common denominator (l.c.d.) is #(e^x+2)^3#. Thus,
#(d^2y)/(dx^2)=((e^x+2)(2e^x+xe^x)+(-2e^x)(2+e^x+xe^x))/(e^x+2)^3#

Distributing the factors, we get the following, which allows us to operate some cancelling, as I will remark in collors:

#(d^2y)/(dx^2)=(color(red)(2e^(2x))color(green)(+xe^(2x))+color(blue)(4e^x)+2xe^xcolor(blue)(-4e^x)color(red)(-2e^(2x))color(green)(-2)xe^(2x))/(e^x+2)^3#

Final simplification:

#(d^2y)/(dx^2)=(2xe^x-xe^(2x))/(e^x+2)^3=color(green)((e^x(2-e^x)x)/(e^x+2)^3)#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the second derivative of ( \frac{x}{2+ e^x} ), first differentiate it once using the quotient rule, then differentiate the result again using the quotient rule and chain rule.

First derivative: [ \frac{d}{dx} \left( \frac{x}{2+ e^x} \right) = \frac{(2+e^x)(1) - (x)(e^x)}{(2+e^x)^2} ]

Second derivative: [ \frac{d^2}{dx^2} \left( \frac{x}{2+ e^x} \right) = \frac{(2+e^x)(0) - (1)(e^x) - (x)(e^x)}{(2+e^x)^2} - \frac{(2+e^x)^2(e^x) - 2(2+e^x)(1)(e^x)}{(2+e^x)^4} ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find the second derivative of ( \frac{x}{2 + e^x} ), you first need to find the first derivative, then differentiate it again.

( f(x) = \frac{x}{2 + e^x} )

First derivative, ( f'(x) = \frac{(2 + e^x)(1) - (x)(0 + e^x)}{(2 + e^x)^2} )

( f'(x) = \frac{2 + e^x - xe^x}{(2 + e^x)^2} )

Now, find the second derivative by differentiating ( f'(x) ).

( f''(x) = \frac{((2 + e^x)^2)(0 + e^x) - (2 + e^x - xe^x)(2e^x)}{(2 + e^x)^4} )

( f''(x) = \frac{(2 + e^x)(e^x) - (2e^x + e^{2x} - 2xe^x)}{(2 + e^x)^3} )

( f''(x) = \frac{2e^x + e^{2x} - 2xe^x - 2e^x}{(2 + e^x)^3} )

( f''(x) = \frac{e^x + e^{2x} - 2xe^x}{(2 + e^x)^3} )

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
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.

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.

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7