Calculating Basic Partial Derivatives

Question

I know this is basic and I've read up on partial derivatives but to double check and make sure I have it correct. I have three problems where I need to calculate the partial derivative with respect to X. Z is another variable. C is constant but unknown:

1. Y = Z*X*5
2. Y = XC + ZC + $15x^{2}$
3. y = $X^{1/2}$ + Z

My answers are as follows:

1. 5Z (is it 5Z or just 5?)
2. Z + 15
3. Z(or is it $.5X^{-1/2}$)

Any and all help would be greatly appreciated. If my answers are wrong, please provide the correct one so I can work backwards to solve the problem. Thank you.

Answer 1

For #2, consider the function $f(x)=xc$, where $c$ is a constant. Surely you agree that $f'(x)=c$, not $f'(x)=0$. Your calculation for 2 is quite off.

Assuming $x$ and $X$ are the same variable, the answer to 2 should be $\frac{\partial}{\partial X} Y = C+30X$.

Answer 2

Well, you have the functions: $$y=5xz \tag{1}$$ $$y=cx+cz+15x^2 \tag{2}$$ $$y=x^{1/2}+z \tag{3}$$ Perhaps it would help to solve the problems if you take a look at Wikipedia's description of a partial derivative:

In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary).

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Your solution to the first one is correct. Since we are taking partial derivatives with respect to $x$, we consider $5z$ as a constant. Therefore: $$\frac{\partial}{\partial x}(5xz)=5z\cdot \frac{\partial}{\partial x}(x)=5z$$

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The second one is incorrect. We consider $cz$ as a constant since it does not contain $x$. Recall that the derivative of a constant is $0$. $$\frac{\partial}{\partial x}(cx+cz+15x^2)=\frac{\partial}{\partial x}(cx)+\color{red}{\frac{\partial}{\partial x}(cz)}+\frac{\partial}{\partial x}(15x^2)$$ $$=c\cdot \frac{\partial}{\partial x}(x)+15\cdot \frac{\partial}{\partial x}(x^2)=c+30x$$

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Do you think you can do the third one?