MHB 2.2.1 AP Calculus Exam .... derivative with ln

[karush]

If $f(x)=7x-3+\ln(x),$ then $f'(1)=$
$a.4\quad b. 5\quad c. 6\quad d. 7\quad e. 8$

see if you can solve this before see the proposed solution

Spoiler

since
$$f'(x)=7+\dfrac{1}{x}$$
so then
$$f'(x)=7+\dfrac{1}{1}=7+1=8\textit{ is (e)}$$

 

[skeeter]

Spoiler

I "see" $7 + \dfrac{1}{x}$ as the derivative ...

here's one for you ...

 

[karush]

Ok we can start with $g'(3)$ and the inverse

Spoiler

$g'(3)=\dfrac{1}{f'(g(3))}$
The rest is just plug in since $g(3)=f^{-1}(3)$
Then from $f(6)=3$ we obtain $f'(3)=6=g(3)$
So now we have
$g'(3)=\dfrac{1}{f'(6)}$
And it was given that $f'(6)=-2$ so we have
$g'(3)=-\dfrac{1}{2}$

to beta pdf
https://documentcloud.adobe.com/link/track?uri=urn:aaid:scds:US:7554449c-33e9-4290-8c9f-dcae56854fc5

 

[Monoxdifly]

I will go with karush's question instead since my brain is closer to that level of question.

Spoiler

$$f'(1)=7+\frac11=7+1=8$$
The answer is D.

 

[karush]

I will go with karush's question instead since my brain is closer to that level of question.

Spoiler

$$f'(1)=7+\frac11=7+1=8$$
The answer is D.

D is 7

 

[Monoxdifly]

D is 7

Oops, sorry. I mean E.

Duh, even though I went with the easy question I still ended up misread the options. Sometimes I think I have ADD.

 

[karush]

Well you are probably
Better at math than I am

 

[Monoxdifly]

Well you are probably
Better at math than I am

Nah. Otherwise, I would do skeeter's question instead.

 

[HallsofIvy]

For skeeter's question use the fact that \frac{df^{-1}}{dx}= \frac{1}{\frac{df}{dx}$. We are asked to find g'(3) wheng(x)= f^{-1}{x)[/tex]. We know that g(f(x))= x and see that f(6)= 3. Since f'(6)= -2, g'(3)= -1/2, "(A)".

Actually, the fact that the only derivative we are given f'(6)= -2 is pretty much a "give away"!