MHB Function differentiation, show that....

[Petrus]

Hello MHB,
Sorry for the bad title as I did not know what to name this but this is a problem from my calculus exam which I have not decide if I shall travel 2h to get my exam and see if I got some less point then I should.. (I just got the facit for the exam and I think that I am between one higher grade that I should get)

"Supposed that $$f(x)$$ is a function so that a tangent to function graf is $$x=2$$ goes through $$(0,1)$$. Supposed that also $$f(2)=5$$ and $$g(x)=\frac{f(x)}{x}$$

(a) show that $$g'(2)=-\frac{1}{4}$$

What I did:
$$g'(x)=\frac{f'(x)x-f(x)}{x^2}$$
We know that $$g'(2)=-\frac{1}{4}$$ and $$f(2)=5$$ so I did do like this
$$-\frac{1}{4}=\frac{2f'(x)-5}{2^2}$$ that means $$f'(2)=2$$ and that means this is true only if $$f'(2)=2$$

Facit say:
they first start to calculate the tangent to the graph of $$f$$ for $$x=2 $$ gives of $$f'(2)=\frac{y-f(2)}{x-2}$$

we know that $$f(2)=5$$ also that the point $$(x,y)=(0,1)$$ and if we put those value on the above we get that $$f'(2)=2$$ and then they did same as me but what so you think about how I did it?

Regards,
$$|\pi\rangle$$

 

[M R]

Re: Function derivate, show that

Hi Petrus,

You got it the wrong way around. (Tongueout)

Work out f'(2). Work out g'(x). Use those to get g'(2)=-1/4. :)

 

[MarkFL]

Hello MHB,
Sorry for the bad title as I did not know what to name this...

Hey Petrus,

In English, the act of computing derivatives is referred to as differentiation. You might say, "I will now differentiate this function with respect to the independent variable." Or, "Differentiation of the following function with respect to $x$ gives us..."