Differentiation help please!

Hi, It’s been a while since I done calculus & I’d really appreciate someone checking if I’ve done this calculation right.

I have D= (-q)/(4πa ) 1/(1-R)

R=r/a Where r=(x^2+y^2+z^2 )^(1/2)

I want to find (∂D)/∂x

I’ve worked it out to be (-q)/4π x/(r(1-R)) is this correct?

Thanks

Reply 1

Clarity Incognito

Cheese Toasty

In the question,

D = (\frac{-q}{4na}) (\frac{1}{1-R})

right?

Reply 2

Cheese Toasty

Yes that's right. (Sorry I couldn't work out how to write equations properly on here)

Reply 3

Clarity Incognito

Cheese Toasty

Yes that's right. (Sorry I couldn't work out how to write equations properly on here)

No that's fine, have a look at this if you intend to use the maths forum further: http://www.thestudentroom.co.uk/wiki/LaTex

Here's what I got for the differentiation. It would be useful if you could show your method too.

D = \left( (\dfrac{-q}{4na})(\dfrac{1}{1-R}) \right)

(\dfrac{\partial D}{\partial R})_{q,n,a} = (\dfrac{-q}{4na})(\dfrac{1}{(1-R)^2})

R = \dfrac{r}{a}

(\dfrac{\partial R}{\partial r})_{a} = \dfrac{1}{a}

r =(x^2 + y^2 + z^2)^{\frac{1}{2}}

(\dfrac{\partial r}{\partial x})_{y,z} = x(x^2 + y^2 + z^2)^{-\frac{1}{2}} = \dfrac{x}{r}

\dfrac{\partial D}{\partial x} = (\dfrac{\partial D}{\partial R}) (\dfrac{\partial R}{\partial r}) (\dfrac{\partial r}{\partial x})

\dfrac{\partial D}{\partial x} =(\dfrac{-q}{4na^2})(\dfrac{x}{r(1-R)^2})

Reply 4

Cheese Toasty

Thanks, heres my attempt at explaining what I did using Latex.

Since everything in the first term is a constant I only differentiated the second term. (I don't think I made clear the fact that a is a constant in my original post).

\displaystyle \frac{\partial}{\partial x}\frac{1}{1-R}

Using reciprocal rule (from quotient rule)

\displaystyle \frac{-g'(x)}{(g(x))^2}

And using chain rule

(g'(x))= \displaystyle \frac{\partial}{\partial u}1-\frac{u^\frac{1}{2}}{a}

\displaystyle \frac{\partial}{\partial x}u

With

(x^2+y^2+z^2)=u

\displaystyle \frac{\partial}{\partial u}1-\frac{u^\frac{1}{2}}{a}

-\frac{1}{2}\frac{u^-\frac{1}{2}}{a}

\displaystyle \frac{\partial}{\partial x}u

2x

Therefore

(-g'(x))= \frac{ax}{u^\frac{1}{2}}

=\frac{ax}{r}

and

\displaystyle \frac{-g'(x)}{(g(x))^2}

\frac{ax}{r(1-R)}

\displaystyle \frac{\partial}{\partial x}=\frac{-q}{4n} . \frac{x}{r(1-R)}

Reply 5

Clarity Incognito

Cheese Toasty

\displaystyle \frac{\partial}{\partial x}\frac{1}{1-R}

Using reciprocal rule (from quotient rule)

\displaystyle \frac{-g'(x)}{(g(x))^2}

And using chain rule

(g'(x))= \displaystyle \frac{\partial}{\partial u}1-\frac{u^\frac{1}{2}}{a}

\displaystyle \frac{\partial}{\partial x}u

With

(x^2+y^2+z^2)=u

\displaystyle \frac{\partial}{\partial u}1-\frac{u^\frac{1}{2}}{a}

-\frac{1}{2}\frac{u^-\frac{1}{2}}{a}

\displaystyle \frac{\partial}{\partial x}u

2x

Therefore

(-g'(x))= \frac{ax}{u^\frac{1}{2}}

=\frac{ax}{r}

and

\displaystyle \frac{-g'(x)}{(g(x))^2}

\frac{ax}{r(1-R)}

\displaystyle \frac{\partial}{\partial x}=\frac{-q}{4n} . \frac{x}{r(1-R)}

Yeah, that was a lot more useful to analyse, look at

(g(x))^2

when you plug it back in.

Also, these lines do not follow. Try spotting the mistake.

\displaystyle \frac{\partial}{\partial u}1-\frac{u^\frac{1}{2}}{a}

-\frac{1}{2}\frac{u^-\frac{1}{2}}{a}

\displaystyle \frac{\partial}{\partial x}u

2x

Therefore

(-g'(x))= \frac{ax}{u^\frac{1}{2}}

=\frac{ax}{r}

Reply 6

Cheese Toasty

Thanks for your patience. Think I see what I've done wrong (apart from forgetting to square (1-R)) I think I substituted r for u to soon. I should have got:

(-g'(x))= \frac{xu^-\frac{1}{2}}{a}

Then

\displaystyle \frac{-g'(x)}{(g(x))^2}

\frac{a}{xu^-\frac{1}{2}}

\frac{1}{(1-R)^2}

\frac{a}{\frac{x}{r}}

\frac{1}{(1-R)^2}

\frac{x}{ar(1-R)^2}

\displaystyle \frac{\partial}{\partial x}D=\frac{-q}{4na^2} . \frac{x}{r(1-R)}

Reply 7

Clarity Incognito

Cheese Toasty

Thanks for your patience. Think I see what I've done wrong (apart from forgetting to square (1-R)) I think I substituted r for u to soon. I should have got: