Partial Differentiation

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April 25th 2010, 06:01 AM
matlabnoob

Partial Differentiation

hi all,
can anyone help me with this?how do you get ∂A/∂q ? i amv.v. confused(Speechless)
i dont need all details or workings,just shorthand way of seeing how to get ∂A/∂q. how do ppl do this quickly?

A = q [(a – qsinѲ)secѲ + (b – qcosѲ)cosecѲ]
∂A/∂q = (a – qsinѲ)secѲ + (b – qcosѲ) – q(sinѲsecѲ + cosѲcosecѲ)

thanks!!
April 25th 2010, 08:21 AM
HallsofIvy

Quote:

Originally Posted by matlabnoob

hi all,
can anyone help me with this?how do you get ∂A/∂q ? i amv.v. confused(Speechless)
i dont need all details or workings,just shorthand way of seeing how to get ∂A/∂q. how do ppl do this quickly?

A = q [(a – qsinѲ)secѲ + (b – qcosѲ)cosecѲ]
∂A/∂q = (a – qsinѲ)secѲ + (b – qcosѲ) – q(sinѲsecѲ + cosѲcosecѲ)

thanks!!

One way to do this would be to use the product rule. The derivative of "q" is 1, of course, and the derivative of $(a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta)$ is $-sin(\theta)sec(\theta)- cos(\theta)cosec(\theta)$ because the derivative of "Cq" where C is not a function of q, is just C itself.

Putting those together, using the product rule:
$(q)'(a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta))+ (q)(a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta))'$
$= a- qsin\theta)sec(\theta)+ (b- qcos(\theta))cosec(\theta)+ q(sin(\theta)sec(\theta)- cos(\theta)cosec(\theta))$

Another way to do this is to go ahead and multiply that out:
$aq sec(\theta)- q^2 sin(\theta)sec(\theta)+ bq cosec(\theta)- q^2 cos(\theta)cosec(\theta)$ .

Now, the derivative of Cq is just C and the derivative of $Cq^2$ is 2Cq so the derivative of that is
$a sec(\theta)- 2q sin(\theta)sec(\theta)+ b cosec(\theta)- 2q cos(\theta)cosec(\theta)$ .

Those two methods should give you the same thing.
April 25th 2010, 08:25 AM
Tikoloshe

Suppose A = q [(5 – 3q)*2 + (11 – 7q)*13]. I assume you can find $\tfrac{dA}{dq}$ rather “quickly” as you say (just several applications of the product rule).

Partial differentiation is just like the above. Any term which involves a variable which is not the variable you’re considering can be treated as a constant (e.g., when finding $\tfrac{\partial A}{\partial q}$ you can treat anything involving $a$ or $\theta$ as numbers). I hope this has clarified and not exacerbated your confusion.

P.S.
I think you are missing a factor of $\csc\theta$ in the second term of your answer.
April 25th 2010, 09:32 AM
matlabnoob

aha..(Happy)thank you both of you!