Partial Derivative: Finding t with Respect to x | Step-by-Step Guide

[galipop]

Hi All,

Can someone refresh my memory and show me how to find the following partial derivate:

t=\frac{x}{\sqrt{x^2+y^2}}

with respect to x.

Thanks

 

[arildno]

The partial derivative of t(x,y) with respect to x is found by differentiating the expression as in 1-variable calculus; just remember that y is to be considered a constant. For example:
t(x,y)=xy\rightarrow\frac{\partial{t}}{\partial{x}}=y

 

[galipop]

Yeah I understand the concept. It's just that I've forgotten how to find the derivative of a quotient. The textbook that I have doesn't explain it clearly.

 

[arildno]

The simplest way is to solve for the derivative of t w.r. to x by differentiating the equivalent equation with respect to x:
\sqrt{x^{2}+y^{2}}t(x,y)=x

Another easy way is (I'll give it for one variable):
t(x)=\frac{f(x)}{g(x)}
Use the product rule and the chain rule:
\frac{dt}{dx}=\frac{f'(x)}{g(x)}-\frac{f(x)}{g^{2}(x)}*g'(x)

 

[TALewis]

I remember the quotient rule as "Lo d Hi minus Hi d Lo over Lo squared," or:

\frac{d}{dx}\,\frac{f(x)}{g(x)}=\frac{gf' - fg'}{g^2}

 

[NateTG]

Personally I generally use:
\frac{a}{b}=a (b)^{-1}
so in your case
\frac{x}{\sqrt{x^2+y^2}}=x (x^2+y^2)^{-\frac{1}{2}}

 

[e(ho0n3]

Yeah I understand the concept. It's just that I've forgotten how to find the derivative of a quotient. The textbook that I have doesn't explain it clearly.

When I forget how do calculate a particular derivate I always derive it by using the definition of the derivate, i.e.

\frac{d}{dx}f(x) = \lim_{h \to \infty} \frac{f(x+h)-f(x)}{h}​

e(ho0n3

 

[galipop]

Personally I generally use:
\frac{a}{b}=a (b)^{-1}
so in your case
\frac{x}{\sqrt{x^2+y^2}}=x (x^2+y^2)^{-\frac{1}{2}}

How do you proceed using this form?

 

[cookiemonster]

Product rule and power rule.

f'*g + g'*f, where f = x and g = (x^2 + y^2)^(-1/2), so f' = 1 and g' = -x(x^2 + y^2)^(-3/2)

And e(ho0n3, don't you mean limit as h goes to 0?

cookiemonster

 

[galipop]

Product rule and power rule.

f'*g + g'*f, where f = x and g = (x^2 + y^2)^(-1/2), so f' = 1 and g' = -x(x^2 + y^2)^(-3/2)

And e(ho0n3, don't you mean limit as h goes to 0?

cookiemonster

Thanks Cookiemonster.