Question

Question #147ce

Answer 1

using the product rule and the rules for `ln` and `e`,

`d/dxln(xe^(-2x))=(1-2x)/x`

Explanation:

alright so it's been a couple years since i've done this but let's see if i can remember.
so we're gonna use the following rules:

`d/dxln(u) = (d/dx(u))/u`

`d/dxe^u=(d/dx(u))e^u`

`d/dxax=a`

`d/dxuv=ud/dx(v)+vd/dx(u)`

where `u` and `v` are any differentiable expressions, `u'` and `v'` are their derivatives, and `a` is any real number.

so taking it step by step, we start here:
1.`d/dx ln(xe^(-2x))`

now we can express that as `ln(u)` where `u=xe^(-2x)`, so if we can derive `u`, we just put another `u` under that and we'll be done.
2.`d/dx ln(xe^(-2x)) = (d/dx(xe^-2x))/(xe^-2x)`

now to find `d/dx(xe^-2x)`, or `u'`, we start by applying the product rule (letting `v=x` and `u=e^(-2x)`). note that `ud/dx(v)` becomes `u` when `v=x`, since `d/dxx=1`

3.`d/dx(xe^-2x) = e^(-2x)+xd/dxe^(-2x)`

now we just need to find the derivative of `e^(-2x)` and we can wrap it all up!

4.`d/dx(e^(-2x)) = e^(-2x)d/dx(-2x)`

the derivative of `-2x` is simply `-2`, so this brings us to:

5.`d/dx(e^(-2x)) = -2e^(-2x)`

we can then plug this into step (3), giving:

3.`d/dx(xe^(-2x)) = e^(-2x)-2xe^(-2x)`

Then we plug that into (2):

2.`d/dx ln(xe^(-2x)) = (e^(-2x) - 2xe^(-2x))/(xe^(-2))`

`d/dxln(xe^(-2x))=(e^(-2x)(1-2x))/(xe^(-2x))`

`d/dxln(xe^(-2x))=(1-2x)/x`

and we've solved it!
(this is my first time posting an answer on this site so please let me know if you have any questions or if anything was unclear

Answer 2

`dy/dx=(1-2x)/x`

Explanation:

The function is `y=ln(x*e^(-2x))`

Let's differentiate both sides with respect to `x` using the chain rule and the product rule.

The chain rule states that `->`

`d/dx(V(U(x)))=d/dxV(U(x)) xx d/dx(U(x)) xx d/dx(x)`

Whenever some functions are nested inside other functions (example `->` `e^(x^2+5x)` or `sin(12x+x^5)` or `ln(10x+2x^2+3x^3)`), we use the chain rule to differentiate them.

The product rule states that `->`

`d/dx(V(x)*U(x))=V(x)d/dxU(x)+U(x)d/dxV(x)`

This rule is used when we we have a function where `n` number of functions are in multiplication.

Now back to our function `y=ln(x*e^(-2x))`

`d/dxy=d/dx(ln(x*e^(-2x)))`

`dy/dx=1/(x*e^(-2x)) xx d/dx(x*e^(-2x))`

`dy/dx=1/(x*e^(-2x)) xx (e^(-2x)*d/dx(x) + x*d/dx(e^(-2x)) )`

`dy/dx=1/(x*e^(-2x)) xx (e^(-2x)+e^(-2x)*x*d/dx(-2x)*)`

`dy/dx=1/(x*e^(-2x)) xx (e^(-2x)+(-2)*x*e^(-2x))`

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

Now we can take `e^(-2x)` common from the numerator.

`dy/dx=(e^(-2x)(1+(-2)*x))/(x*e^(-2x))`

`dy/dx=(cancele^(-2x)(1+(-2)*x))/(x*cancele^(-2x))`

`dy/dx=(1-2x)/x`