Partial Derivatives Example 8


Uploaded by TheIntegralCALC on 19.07.2010

Transcript:
Hi everyone. Welcome back to integralCALC.com. WeÕre going to be doing another partial derivatives
problem today. This one asks us to find the first-order partial derivatives of this function
here, function Q and because we have two variables in the problem, we are finding partial derivatives
first with respect to K and then with respect to L. So the notation for that is the following.
The partial derivative of the function Q with respect to K and the partial derivative of
Q with respect to L and this problem is fairly simple. All we need to do is use the power
rule because when weÕre taking the derivative with respect to K, L weÕre treating as a
constant. So if we plug in a number for L like 2, what do we have? 2 to the 0.6, right?
So this 2 to the 0.6 would simplify as 1.51 something, right? This whole L to the 0.6
becomes just a number and it would get absorbed into the coefficient. It would be multiplied
by 10 and you just have 10 K to the 0.4 and then we would take the derivative with respect
to K.
So like I said, weÕre treating the other variable as a constant and in the case of
this function, thatÕs like a variable. Since everything is multiplied together, it will
just get absorbed into the coefficient. So the derivative here, we will multiply the
exponent on K by the coefficient here. 10 times 0.4 is simply 4 and then we subtract
1 from the exponent. 0.4 minus 1 is negative 0.6 and then the L we leave alone because
remember it was absorbed into the coefficient when we treat it like a constant. So itÕs
just L to the 0.6 there.
Then when we take the derivative with respect to L, we do the same thing. We treat K as
a constant. If we plugged in the number for K, this whole thing would just turn into a
number and it would get absorbed here with this 10 so this whole 10 K to the 0.4 acts
like the coefficient on the L.
So all that is going to stay and all we need to do is multiply the exponent here on the
L by 10 out in front. So we will get 6 and then we leave K to the 0.4 alone and then
L, we subtract 1 from the exponent. 0.6 minus 1 is negative 0.4 and thatÕs it.
Those are our first-order partial derivatives and these are your final answers. Thanks guys.
See you next time.
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