Evaluate the derivative of the function at the given point.?Use a graphing utility to verify your result. (If an answer is undefined, enter UNDEFINED. Round your answer to three decimal...

Evaluate the derivative of the function at the given point.?

Use a graphing utility to verify your result. (If an answer is undefined, enter UNDEFINED. Round your answer to three decimal places.)

`f(x)=3/(x^2-4x)^2`  at `(5,3/25)`  

Asked on by tagnaouti

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tiburtius's profile pic

tiburtius | High School Teacher | (Level 2) Educator

Posted on

Chain rule:

`(f(g(x)))'=f'(g(x))g'(x)`

Sum rule:

`(f(x)pm g(x))'=f'(x)pm g'(x)`

So by using the above rules we can differentiate our function, but first let's rewrite function `f` to get more convenient form.

`f(x)=3/(x^2-4x)^2=3(x^2-4x)^(-2)`

`f'(x)=3cdot(-2)(x^2-4x)^(-3)cdot(2x-4)=-(6(2x-4))/(x^2-4x)^3`

So to calculate `f'(5)` you simply put 5 instead of `x.`

`f'(5)=-36/125=-0,288` 

I have to point out that this is not rounded result, that is there are only 3 decimal places, no more.

Now to be able to graph the tangent at point `(5,3/25)`  we must find equation of the tangent line which is given by formula:

`y=f'(x)x+l`                                                                       (1)

Now to calculate `l` we simpliput all data we have into formula (1).

`3/25=-36/125cdot5+l`

`3/25=-36/25+l`

`l=39/25=1.56`

Blue line represents your function `f` while red line represents tangent at point `(5,3/25)`.

lfryerda's profile pic

lfryerda | High School Teacher | (Level 2) Educator

Posted on

To evaluate the derivative, it is best to rearrange the function slightly to make it easier to take the derivative.

`f(x)=3/(x^2-4x)^2`

`=3(x^2-4x)^{-2}`

Now using the power rule and chain rule:

`f'(x)=3(-2)(x^2-4x)^{-3}(2x-4)`  and evaluate at x=5

`f'(5)=-6(25-20)^{-3}(10-4)`

`=-36/125`

The derivative at x=5 is `f'(5)=-36/125` .

The graph and its derivative is:

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