derivative of a function

[edwindelasacha]

Having problems with this one.. I have the question and final answer but dont know how to get to the answer..

Q) N=2(x-L/2)(x-L)/L^2
A) dN/dx = 4x/L^2 - 3/L

Any suggestions please?

 

[Quaid]

dont know how to get to the answer

Q) N=2(x-L/2)(x-L)/L^2

A) dN/dx = 4x/L^2 - 3/L

Hi:

Your statement in red above is vague, so I'll guess at what you already understand and why you're stuck.

Start by multiplying everything in the expression for function N; this will yield a quadratic polynomial.

Use the Power Rule, term-by-term, to arrive at dN/dx

If you're still stuck, please be specific and show your efforts.

Cheers :cool:

 

[JeffM]

Having problems with this one.. I have the question and final answer but dont know how to get to the answer..

Q) N=2(x-L/2)(x-L)/L^2
A) dN/dx = 4x/L^2 - 3/L

Any suggestions please?

$\displaystyle N = \dfrac{2(x - 0.5L)(x - L)}{L^2}.$ Is that correct? The most direct (but not the only) approach is to simplify before differentiating.

So $\displaystyle N = \dfrac{(2x - L)(x - L)}{L^2} = \dfrac{2x^2 - 3Lx + L^2}{L^2} = \left(\dfrac{2}{L^2}\right)x^2 + \left(\dfrac{-3}{L}\right)x + 1.$

Now what?

 

[edwindelasacha]

$\displaystyle N = \dfrac{2(x - 0.5L)(x - L)}{L^2}.$ Is that correct? The most direct (but not the only) approach is to simplify before differentiating.

So $\displaystyle N = \dfrac{(2x - L)(x - L)}{L^2} = \dfrac{2x^2 - 3Lx + L^2}{L^2} = \left(\dfrac{2}{L^2}\right)x^2 + \left(\dfrac{-3}{L}\right)x + 1.$

Now what?

Yes that is correct. I put my problem into an online solver to find the derivative and it output this result: 4x-3L/L^2 which is what I was looking for.

Thanks

 

[Quaid]

I put my problem into an online solver to find the derivative

This is the suggestion that you were asking for? :???:

it output this result: 4x-3L/L^2

Oops, be careful. Without grouping symbols around the numerator, your typing means this:

$\displaystyle 4x - \dfrac{3L}{L^2}$

You do not want to forget required grouping symbols, especially when entering expressions into your on-line solver during an exam! :lol: