Find the numerical value of the derivatives

[qwert]

The question goes like this:

Find the numerical value of the derivatives of the following curve at the given point,

when x= -2

thank you very much!


I tried 2 ways to do this question:

1) Chain rule

sub x= -2

= 3.927

2) differentiating

like i would differentiate

(which means without the brackets)

sub x=-2

= 0.27


Both answers are wrong. Right answer is 0.135

Thank you very much!

 

[mathmari]

\(\displaystyle \frac{dy}{dx}=-e^{x}(1+x) \)
sub x=-2
= \(\displaystyle e^{-2}=0.135 \)

 

[qwert]

\(\displaystyle \frac{dy}{dx}=-e^{x}(1+x) \)
sub x=-2
= \(\displaystyle e^{-2}=0.135 \)

Hi! Firstly i'd like to thank you for the help!
But happens to the x in xe^x in your working? thank you! im a little confused

 

[mathmari]

\(\displaystyle (4-xe^{x})'=(4)'-(xe^{x})'=0-[(x)'e^{x}+x(e^{x})']=-[e^{x}+xe^{x}]=-e^{x}(1+x) \)

 

[JeffM]

The question goes like this:

Find the numerical value of the derivatives of the following curve at the given point,

when x= -2

thank you very much!


I tried 2 ways to do this question:

1) Chain rule

sub x= -2

= 3.927

2) differentiating

like i would differentiate

(which means without the brackets)

sub x=-2

= 0.27


Both answers are wrong. Right answer is 0.135

Thank you very much!

Neither derivative is correct. In fact, the first method is so far off the track that I cannot even guess what you were thinking. (Don't explain because an explanation will merely reinforce error.)

\(\displaystyle Let\ w = xe^x \implies y = 4 - xe^x = 4 - w \implies \dfrac{dy}{dw} = - 1.\)

\(\displaystyle Let\ w = uv\ and\ u = x\ and\ v = e^x \implies \dfrac{du}{dx} = 1\ and \dfrac{dv}{dx} = e^x\ and \dfrac{dw}{dx} = u * \dfrac{dv}{dx} + v * \dfrac{du}{dx} = x * e^x + e^x * 1 = e^x(x + 1).\) Multiplication Rule.

\(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{dw} * \dfrac{dw}{dx} = -1 * e^x(x + 1) = -e^x * (x + 1).\) Chain Rule.

\(\displaystyle - e^{-2} *(-2 + 1) = - \dfrac{1}{e^2} * (-1) = \dfrac{1}{e^2} \approx 0.135.\)

 

[HallsofIvy]

The question goes like this:

Find the numerical value of the derivatives of the following curve at the given point,

when x= -2

thank you very much!


I tried 2 ways to do this question:

1) Chain rule

You appear to be thinking of this as \(\displaystyle y= (4- xe^x)^1\) and using the power rule: \(\displaystyle (x^n)'= nx^{n-1}\). But with n= 1, n-1= 0, NOT -2! Of course, then \(\displaystyle (1)(4- xe^x)^0(4- xe^x)'\) is just \(\displaystyle (1)(1)(4- xe^x)'\) so there really is no point in doing that! To differentiate \(\displaystyle 4- xe^x\) itself, the derivative of "4" is, of course, 0 and you have to use the product rule on \(\displaystyle -xe^x\): \(\displaystyle (xe^x)'=(-x)'e^x- x(e^x)'= -e^x- xe^x\), not just \(\displaystyle -xe^x\).

sub x= -2

= 3.927

2) differentiating

like i would differentiate

(which means without the brackets)

Again, as above, the derivative of \(\displaystyle -xe^x\) is \(\displaystyle -e^x- xe^x\). You appear to be treating that first "x" as if it were an constant- and it isn't!

sub x=-2

= 0.27


Both answers are wrong. Right answer is 0.135

Thank you very much!