Basic partial differentials

[nz.lauren]

Hey, I'm brand new to calculus, so I'm probably missing something really obvious, but my textbook's answers don't show the working so I'm totally lost!

The question is: Find all the partial derivatives and the total differential of y=3x^1/3 z^2/3. The answer for the partial derivatives is f_x(x,z) = x^-2/3 z^2/3, f_z(x,z) = 2x^1/3 z^-1/3.

Basically I understand for f_x(x,z) that the 'z^2/3' is treated as a constant, but I don't understand what happens to the constant '3' in 3x^1/3. I thought that the multiplication by a constant rule would mean that the derivative of 3x^1/3 would be 3(1x^-2/3) which is 3x^-2/3. Even if I took the derivative of the 3 seperately, wouldn't this mean it becomes 0 * x^-2/3 equalling 0?

Also for f_z(x,z), I thought I should treat only x^1/3 as the constant, meaning I multiply the derivative of 3 (which is 0?) by x^1/3 z^-1/3, giving me 0? Am I wrong in treating the 3 in '3x^1/3' as a constant? How does the 3 become a 2?!

Hope this makes sense!

 

[galactus]

Find all the partial derivatives and the total differential of y=3x^1/3 z^2/3. The answer for the partial derivatives is f_x(x,z) = x^-2/3 z^2/3, f_z(x,z) = 2x^1/3 z^-1/3.

but I don't understand what happens to the constant '3' in 3x^1/3

.

When you take the derivative of \(\displaystyle \displaystyle 3x^{\frac{1}{3}}\). You multiply the exponent by the constant. Thus, \(\displaystyle \displaystyle 3\cdot \frac{1}{3}=1\). That's what happened to the 3.
Then, subtract 1 from the exponent as you done. Getting, \(\displaystyle \displaystyle x^{\frac{-2}{3}}=\frac{1}{x^{\frac{2}{3}}}\). Since the z term is a constant, it gets multiplied by the x term we now have. Ultimately getting \(\displaystyle \displaystyle f_{x}=\frac{z^{\frac{2}{3}}}{x^{\frac{2}{3}}}\)

Also for f_z(x,z), I thought I should treat only x^1/3 as the constant, meaning I multiply the derivative of 3 (which is 0?) by x^1/3 z^-1/3, giving me 0? Am I wrong in treating the 3 in '3x^1/3' as a constant? How does the 3 become a 2?!

Same idea as above only the x term is the constant now. By multiplying 2/3 by 3. Which gives 2. Then, subtract 1 from 2/3 and get -1/3 for the exponent.

Differentiate the z term and get \(\displaystyle \displaystyle \frac{2}{3}z^{\frac{-1}{3}}=\frac{2}{3z^{\frac{1}{3}}}\).

Now, multiply this by the constant, \(\displaystyle \displaystyle 3x^{\frac{1}{3}}\), and the 3's cancel.

Giving \(\displaystyle \displaystyle f_{z}=\frac{2x^{\frac{1}{3}}}{z^{\frac{1}{3}}}\)

 

[nz.lauren]

Thank you!! I don't know why I was trying to multiply the constant by 1 instead of by the exponent - rookie!!

So when partially differentiating for z, I still multiply the constant (3) by the exponent for z (2/3), even though it is the exponent of z and the constant is in front of x?

I guess that 3x^1/3z^1/3 is equal to 3 * x^1/3 * z^2/3 which is the same as x^1/3 * 3 * z^2/3 right?? So it doesn't matter where the constant is in the equation if its multiplied? Really basic algebra I know but I just want to double check since it's been about 10 years since I last did this!!!

Thanks again!

 

[galactus]

Yep.

 

[mmm4444bot]

It's the Commutative Property of Multiplication that tells us numbers may be multiplied in any order because the resulting product is always the same.

There's a similar property for addition.