Question regarding finding the partial derivative

Question

can someone who is more experienced in partial derivatives explain me this equation? Thank you in advance.

$$f(x,y) = (2x+y)^3\times x$$

1) Calculate the value of $$x=1$$

2) Calculate the value of $$y=2$$

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Task 1:

$$\frac{\partial f}{\partial x} = 3(2x+y)^2 \color{red}{\times 2x + (2x+y)}$$ Can someone please explain from where do we get $$\color{red}{\times 2x + (2x+y)}$$

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Task 2:

$$\frac{\partial f}{\partial y} = 3(2x+y)^2 \color{red}{\times x}$$

Can you please explain why the $$\color{red}{\times x}$$ remains in the equation?

Answer 1

When you take a partial derivative with respect to one variable all work exactly as for the ordinary derivative keeping constant the other variable, for example

$$y=c\implies f(x,c)=g(x)= (2x+c)^3\times x=h(x)\cdot r(x)$$

then

$$\frac{\partial f}{\partial x}=\frac{dg}{dx}=h'(x)\cdot g(x)+h(x)\cdot g'(x)=3\cdot(2x+c)^2\cdot 2\cdot x+(2x+c)^3\cdot 1=\\=6x(2x+y)^2+(2x+y)^3$$

Answer 2

$$\frac{\partial f}{\partial x}=\frac{\partial (2x+y)^3}{\partial (2x+y)}\frac{\partial (2x+y)}{\partial x}x+(2x+y)^3\frac{\partial x}{\partial x}=3(2x+y)^2\times 2x+(2x+y)^3,$$

$$\frac{\partial f}{\partial y}=\frac{\partial (2x+y)^3}{\partial (2x+y)}\frac{\partial (2x+y)}{\partial y}x=3(2x+y)^2\times x.$$

Answer 3

$f(x,y) = x(2x+y)^3$

$$\frac {\partial f}{\partial x}=3(2x+y)^22x+(2x+y)^3=(2x+y)^2(y+8x)$$ $$\frac {\partial f}{\partial y}=3x(2x+y)^2$$