Question

what is the derivative of f(x)=-(x-1)^2-2

WITHOUT

using the chain rule only the product rule or the quotient rule?

Answer 1

why not just use first principles then and ignore all the rules :)

Answer 2

-(x-1)^2-2

-(x^2-2x+1)-2

-x^2 + 2x - 3

now its only power and constant rules ....

Answer 3

Using the product rule:
\[f(x)=-(x-1)(x-1)-2\]
\[f'(x)=-(x-1)(x-1)'+(-(x-1))'(x-1)+(2)'\]

Answer 4

' is prime so take the derivative where you need to and simplify recall that (2)'=0 because the derivative of any constant(number) is 0 with respect to x (or any other variable for that matter).

Answer 5

well what i have is...
f(x)=-(x-1)^2-2
f(x)=-(x-1)-2
f(x)=-x+1-2
But then by my instructors example she has this random example similar to mine...
and so then she obviously has the derivative of f(x) but Spring break just ended and I totally feel retarded because I dont remember how to get the derivative of the f(x) we havent gone over the chain rule so when I look for examples many use the chain rule so IM stuck

Answer 6

you said
**********
well what i have is...
f(x)=-(x-1)^2-2
f(x)=-(x-1)-2
f(x)=-x+1-2
****************

I don't know what that means.
Is this 3 different problems ?

Answer 7

No Phi I believe i answered the problem wrong.

the answer from
f(x)=-(x-1)^2-2

should be
f(x)=x^2+2x-3...
right?

Answer 8

you can expand (x-1)(x-1) to get x^2 -2x +1 (use FOIL)

you can put that into
f(x)= - (x-1)^2-2
don't lose track of the minus sign out front. you get
f(x)= - ( x^2 -2x +1 ) -2
you can distribute the -1 to get
f(x)= -x^2 +2x-1 -2
and combine -1-2 so we get

f(x)= -x^2 +2x-3

Answer 9

But I would learn to do it this way
f(x)=-(x-1)^2-2

let u= (x-1)
then the problem is
\[ f(u) = -u^2 -2 \]
now take the derivative term by term

\[ d f(u)= - d\ u^2 - d\ 2 \]

the derivative of the constant 2 is zero so we only have to worry about the first term

d u^2 = 2 u du
and
\[ d f(u)= -2 u du\]

remember u= x-1
du = d( x-1)
term by term: du = dx - 0= dx
we get (replacing u with (x-1)

\[d(\ -(x-1)^2-2) = -2 (x-1) dx \]
and
\[ \frac{d}{dx} f(x)= -2 (x-1) \]