Differentiation (Product Rule)

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bhaktir

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Differentiation (Product Rule)

z = w^3/2(w + ce^3)


In the above equation, which variables do I treat as constants?

Thanks in advance! :)
 
Tough question. What are you asked to do? Rather crucial information.

If you are asked to find dz/dw, then 'c' is a constant for the calculation.
'e' is probably 'e'. Always a constant. It's bad form to use it as anything else.
If you are asked to find dz/dc, then you won't need the product rule.

Where does that leave us?
 
That's the thing. It's a very vague question. The only thing we're told to do is "differentiate." No other information is given. So, it's pretty confusing: Is c supposed to be treated as the constant or w?

Minor correction: e is to the power of w. Not 3. Does that change anything? :/
 
Fundamental Rule: When in doubt, do it every conceivable way and explain what you did.

Of course, my third comment is rather compelling. If you are supposed to be working on the Product Rule, dz/dc won't do it.
 
Alright, so here's what I'm getting:

dz/dw = w^3/2(1 + wce^w-1) + (w + ce^w)(3/2w^1/2)

[ Original Equation: w^3/2(w + ce^w) ]


Kinda confused.
 
bhaktir said:
Alright, so here's what I'm getting:

dz/dw = w^3/2(1 + wce^w-1) + (w + ce^w)(3/2w^1/2)
ected
[ Original Equation: w^3/2(w + ce^w) ]


Kinda confused.
Click to expand...

Does your function look like:

\(\displaystyle z \ = \ w^{\frac{3}{2}}(w + c \ * \ e^w)\) ..... [edit: typo corrected]

Then it is a wicked one.

\(\displaystyle z \ = \ w^{\frac{5}{2}} \ + \ c \ * \ w^{\frac{3}{2}}\ * \ e^w\)

Hint:

If
\(\displaystyle z_1 \ = \ e^w\)

Then

\(\displaystyle \frac{dz_1}{dw} \ = \ e^w\)
 
Last edited by a moderator:
It's actually,

z = w^3/2(w + ce^w)

There's a c next to e.
--
So, e^w stays as is? Even if c is a constant?
 
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bhaktir said:
It's actually,

z = w^3/2(w + ce^w)

There's a c next to e.

Subhotosh was not trying to ask you if the c should be dropped; that is a typographical error on his part.
Click to expand...

Subhotosh is trying to confirm that the exponent on w is the rational number 3/2 because that is not what you've typed (grouping symbols missing).

In other words, because of the Order of Operations, typing w^3/2 means \(\displaystyle \frac{w^3}{2}\) whereas typing w^(3/2) means \(\displaystyle w^{3/2}\).

Please confirm that your typing is supposed to be the latter, not the former.
 
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bhaktir said:
It's actually,

z = w^3/2(w + ce^w)

There's a c next to e.
--
So, e^w stays as is? Even if c is a constant?
Click to expand...

Yes it does... it is not similar to xn

\(\displaystyle \frac{d}{dx}[x^n] \ = \ n \ * \ x^{n-1}\)

\(\displaystyle \frac{d}{dx}[e^x] \ = \ e^{x}\)

And

\(\displaystyle \frac{d}{dx}[c \ * \ x^n] \ = \ c \ * \ n \ * \ x^{n-1}\)

\(\displaystyle \frac{d}{dx}[c \ * \ e^x] \ = \ c \ * \ e^{x}\)
 
Last edited by a moderator:
The assumption is that you intend \(\displaystyle z=w^{\frac{3}{2}}(w+ce^{w})\)

Without proper grouping symbols, you have written \(\displaystyle z=\frac{w^{3}}{2}(w+ce^{w})\)

The fellas are asking which it is.

You probably are asked to find \(\displaystyle \frac{dz}{dw}\)
 
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