1. ## Differentiate simple question.

Differentiate:

I know that r^6 changes to 6r^5, r changes to 1, and the -1 is eliminated because it turns to 0.. but i'm kind of confused what to do with the - 2/5squareroot of r

Do I have to use the quotient rule? Or is it just bringing 5(r)^1/2 to the numerator, changing it to r^-1/2 and multiplying it by -2.

So (-2)(5)(r)^-1/2
=-10r^-1/2
power rule:
5r^-3/2?

2. You seem to be close. How did the '5' get into the numerator?

3. Originally Posted by kmjt
Differentiate:

I know that r^6 changes to 6r^5, r changes to 1, and the -1 is eliminated because it turns to 0.. but i'm kind of confused what to do with the - 2/5squareroot of r

Do I have to use the quotient rule? Or is it just bringing 5(r)^1/2 to the numerator, changing it to r^-1/2 and multiplying it by -2.

So (-2)(5)(r)^-1/2
=-10r^-1/2
power rule:
5r^-3/2?
let the 2/5 part be as it is. Just find the derivative of 1/sqrt(r)

which will be (-1/2)*r^-3/2

6r^5 + 1/(5r^(3/2))+1

4. Hoow would I write that not as a fraction?

5. Originally Posted by kmjt
Hoow would I write that not as a fraction?
6r^5 + (1/5)*r^-(3/2)+1

Is this what you mean by writing it NOT as a fraction?

6. Originally Posted by harish21
6r^5 + 5*r^-(3/2)+1

Is this what you mean by writing it NOT as a fraction?
5r^-3/2.. so I did do it right in my initial attempt?

7. Originally Posted by kmjt
Differentiate:

I know that r^6 changes to 6r^5, r changes to 1, and the -1 is eliminated because it turns to 0.. but i'm kind of confused what to do with the - 2/5squareroot of r

Do I have to use the quotient rule? Or is it just bringing 5(r)^1/2 to the numerator, changing it to r^-1/2 and multiplying it by -2.

So (-2)(5)(r)^-1/2
=-10r^-1/2
power rule:
5r^-3/2?
the derivative of this part is calculated this way:

-2 /5*-1/2 *r^(-3/2)

Thats where the 2s get cancelled and you are left with

(1/5)*r^(-3/2)

8. Originally Posted by kmjt
5r^-3/2.. so I did do it right in my initial attempt?
I am sorry.. its not 5r^-3/2

Its (1/5) r^-3/2