- #1

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[tex]f(x)=xy , f'(x)=x'y+y'x[/tex]

But is this how to differentiate three terms:

[tex]f(x)=xyz , f'(x)=(x'y+y'x)+(y'z+z'y)+(x'z+z'x)[/tex]

- Thread starter JoshHolloway
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- #1

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[tex]f(x)=xy , f'(x)=x'y+y'x[/tex]

But is this how to differentiate three terms:

[tex]f(x)=xyz , f'(x)=(x'y+y'x)+(y'z+z'y)+(x'z+z'x)[/tex]

- #2

lurflurf

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odd notation you have. I assume you mean y,z are functions of x that is y(x) and z(x). Thoughout ' denotes partial differentiation with respect to x.JoshHolloway said:

[tex]f(x)=xy , f'(x)=x'y+y'x[/tex]

But is this how to differentiate three terms:

[tex]f(x)=xyz , f'(x)=(x'y+y'x)+(y'z+z'y)+(x'z+z'x)[/tex]

anyway

(xyz)'=x'yz+xy'z+zyz'

for a general function f(u1,...,un) you can differentiate holding all functions except one constant at a time. so (In what follows y'=0)

if

g(x)=f(u1(x),u2(x),...,un(x))

g'(x)=(f(u1(x),u2(x),...,un(x)))'

=(f(u1(x),u2(x),...,un(x)))'

=(f(u1(x),u2(y),...,un(y)))'|y=x

+(f(u1(y),u2(x),...,un(y)))'|y=x

+...

+(f(u1(y),u2(y),...,un(x)))'|y=x

or more concreatly for three functions...

if

g(x)=f(u1(x),u2(x),u3(x))

g'(x)=(f(u1(x),u2(x),u3(x)))'

=(f(u1(x),u2(x),u3(x)))'

=(f(u1(x),u2(y),u3(y)))'|y=x

+(f(u1(y),u2(x),u3(y)))'|y=x

+(f(u1(y),u2(y),u3(x)))'|y=x

Last edited:

- #3

rachmaninoff

This is wrong notation, unless[tex]f(x)=xy[/tex]

[tex]f(x,y)=xy[/tex]

and you're differentiating with respect to a third variable,

[tex]\begin{align*} f'(x(t),y(t)) &= \frac{d}{dt} \left[ f(x(t),y(t)) \right] \\

&=\frac{d}{dt}(xy)\\

&=\left( \frac{dx}{dt} \right) y+ \left( \frac{dy}{dt} \right)x\\

&=x'y+y'x \end{align}[/tex]

where your notation refers to the product rule of differentiation - stop me if I'm misreading you.

In which case

[tex]\frac{d}{dt}f(x,y,z)[/tex] where [tex]f(x,y,z)=xyz[/tex]

where

[tex]u=yz,[/tex] (shorthand for u(t) = y(t)z(t) )

[tex]\begin{align*} \frac{d}{dt}f(x,y,z)&=\frac{d}{dt}(xyz)\\

&=\frac{d}{dt}(xu)\\

&=\left( \frac{dx}{dt} \right) u + \left( \frac{du}{dt} \right) x\\

&=x'yz+\frac{d(yz)}{dt}x\end{align}[/tex]

Do you see how this works?

Last edited by a moderator:

- #4

rachmaninoff

[tex]f(t)=x(t) \cdot y(t) \cdot z(t)[/tex]

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- #6

matt grime

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your question is more simply

hod do i differentiate sin(x)cos(x)x^2 with respect to x.

this is straight forward. you can dfferentiate f(x)g(x) so set f(x)=sin(x)cos(x) and g(x)=x^2 and differentiate that.

you should'nt write f(x)=xy at this stage. it's ok to think of y as a function of x, and g(x) as that function, but saying f(x)=xy sort og implies that y is a function of x, though we don't know what (it certainly isn't f(x))

- #7

HallsofIvy

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One way to see it is to write xyz= (xy)z and treat (xy) as a single function of t:

(xyz)'= ((xy)z)'= (xy)'z+ (xy)z' using the "two term" product rule. But (xy)'= x'y+ xy', again by the "two term" product rule. So (xyz)'= (x'y+ xy')z+ (xy)z'= x'yz+ xy'z+ xyz'.

- #8

lurflurf

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f(x) means we provide a number and get one back.JoshHolloway said:

f(x,y) means we provide wto numbes and get one back

When you are differentiating a function where x appears several times like in the product rule f(x)=u(x)v(x) for example here is something that may help. First it would be much easier is all the x's but one were constants. What you can do is let all the x's but one at a time be constants and add up what you get for each.

I will write the x's I consider constant as X and the other one as x to make it more clear.

f(x)=u(x)v(x)

f'(x)=(u(x)v(x))'

=(u(x)v(X))'+(u(X)v(x))'

=u'(x)v(X)+u(X)v'(x)

=u'(x)v(x)+u(x)v'(x)

Nice if you forget the product rule due to head injury.

Also the ever popular

f(x)=x^x

f'(x)=(x^x)'

=(x^X)'+(X^x)'

=X(x^(X-1))+(x^X)log(x)

=x(x^(x-1))+(x^x)log(x)

=(x^x)(1+log(x))

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Excellent. Thanks lurf.

- #10

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So for p(x) = f(x)g(x)h(x) you get [tex]p'\left( x \right) = f'\left( x \right)g\left( x \right)h\left( x \right) + f\left( x \right)g'\left( x \right)h\left( x \right) + f\left( x \right)g\left( x \right)h'\left( x \right)[/tex]. As in, you move the prime 'across one function' for each successive term. Not entirely sure if the method works in general though.

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