What is the slope of #f(x)=-X/e^(x-x^3) # at #x=-1#?

Question

What is the slope of #f(x)=-X/e^(x-x^3)    # at #x=-1#?

Luke Alvarez · Accepted Answer

Slope #=+1#

Given #f(x)=-x/(e^(x-x^3))# at # x=-1#

Determine the first derivative #f' (x)# first, then compute for #f' (-1)# to obtain the slope

We make use of the derivative of quotient for this type of function

use the formula #d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2#

Let #u=x# and #v=e^(x-x^3)#

#f' (x)=d/dx(f(x))=# #d/dx(-x/(e^(x-x^3)))=(-1)*(e^(x-x^3)*d/dx(x)-x*d/dx(e^(x-x^3)))/(e^(x-x^3))^2#

#d/dx(-x/(e^(x-x^3)))=(-1)*((e^(x-x^3)*1-xe^(x-x^3)(1-3x^2)))/(e^(x-x^3))^2#

There is no need for further simplification just use #x=-1# right away

#f' (-1)=(-1)*((e^(-1-(-1)^3)*1-(-1)e^(-1-(-1)^3)(1-3(-1)^2)))/(e^(-1-(-1)^3))^2#

#f' (-1)=(-1)*((e^(-1+1)-(-1)e^(-1+1)(1-3)))/(e^(-1+1))^2#

#f' (-1)=(-1)*((e^(0)-(-1)e^(0)(-2)))/(e^(0))^2#

#f' (-1)=(-1)*((1-2))/(1)^2#

#f' (-1)=+1#

God bless....I hope the explanation is useful.

Ezra Adams · Answer

undefined To find the slope of the function $ f(x) = -\frac{x}{e^{x-x^3}} $ at $ x = -1 $, you need to find the derivative of the function and then evaluate it at $ x = -1 $. The derivative of the function with respect to $ x $ is:

$ f'(x) = -\frac{e^{x-x^3}(1-x^2)}{(e^{x-x^3})^2} - \frac{xe^{x-x^3}(1-3x^2)}{(e^{x-x^3})^2} $

Now, substitute $ x = -1 $ into $ f'(x) $ to find the slope at $ x = -1 $:

$ f'(-1) = -\frac{e^{-1-(-1)^3}(1-(-1)^2)}{(e^{-1-(-1)^3})^2} - \frac{-1e^{-1-(-1)^3}(1-3(-1)^2)}{(e^{-1-(-1)^3})^2} $

$ f'(-1) = -\frac{e^{-1+1}(1-1)}{(e^{-1+1})^2} - \frac{-1e^{-1+1}(1-3)}{(e^{-1+1})^2} $

$ f'(-1) = -\frac{0}{e^0} - \frac{-2e^0}{e^0} $

$ f'(-1) = -\frac{0}{1} - \frac{-2}{1} $

$ f'(-1) = 0 + 2 $

$ f'(-1) = 2 $

So, the slope of $ f(x) = -\frac{x}{e^{x-x^3}} $ at $ x = -1 $ is $ 2 $.