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.

1

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.

1