Question

What is f’(x) F(x)=cos(x)+3x^2+y ?

Answer 1

Assuming `f(x,y) = cos(x)+3x^2+y` then we have:

`(partial f)/(partial x) = 6x -sinx`

Explanation:

There are several issues with this question:

• There is a discrepancy between the use of `f` and `F`. In some cases the use of capitals would describe the function and the corresponding anti-derivative, but I do not believe this is the intended use.

• The function is a dependant upon two variables `x` and `y`, rather than a single variable `x` as given

• As such, we cannot find a derivative `f'(x)`, instead we calculate partial derivatives, unless there is some other relationship between `x` and `y` in terms of some other variable (`t`, say), and the intended use of `f'(x)` is `(df)/dt`, in which case we can use a modified partial derivative version of the chain rule.

Given these discrepancies, I will assume that we seek the partial derivative of`` fwrt `x`, `f_x`, or `(partial f)/(partial x)` for the function defined by:

`f(x,y) = cos(x)+3x^2+y`

Then, we compute the partial derivative of a function of two or more variables by differentiating wrt one variable, whilst the other variables are treated as constant. Thus:

`(partial f)/(partial x) = (partial)/(partial x)(cos(x)+3x^2+y)`

`\ \ \ \ \ \ = (partial)/(partial x)cosx+3(partial)/(partial x)x^2+(partial)/(partial x)y`

`\ \ \ \ \ \ = -sinx + 3(2x) + 0`

`\ \ \ \ \ \ = 6x -sinx`

Should we need it, then we can also compute the partial derivative of `f` wrt `y`, thus:

`(partial f)/(partial y) = (partial)/(partial y)(cos(x)+3x^2+y)`

`\ \ \ \ \ \ = (partial)/(partial y)cosx+3(partial)/(partial y)x^2+(partial)/(partial y)y`

`\ \ \ \ \ \ = 0+0+1`

`\ \ \ \ \ \ = 1`

Finally, if there is a relationship between `x` and `y` in terms of some other variable, `t`, say, then we can use the partial chain rule:

`f'(x) = (df)/dt = (partial f)/(partial x)dx/dt + (partial f)/(partial y)dy/dt`