# Thread: Derivative of arctan in a partial derivative

1. ## Derivative of arctan in a partial derivative

Im working on the derivative f'y(x,y) of the function f(x,y)=arctan(x^2)+ y^3

Because it's a derivative with respect to the variable y, this means that the variable x is held constant, does this mean that the entire arctan(x^2) should be regarded as just a number so the partial derivative equals to 3y^2 in the end??

2. Originally Posted by tinyone
Im working on the derivative f'y(x,y) of the function f(x,y)=arctan(x^2)+ y^3
Because it's a derivative with respect to the variable y, this means that the variable x is held constant, does this mean that the entire arctan(x^2) should be regarded as just a number so the partial derivative equals to 3y^2 in the end??
You are correct.

3. Cheers,

Derivative f'z(x,y,z) = 2(z+3)^4 + ln(xyz)

So keeping y and x constant would give me: f'z= 8(z+3)^3 + (1/xyz) * xy

4. Originally Posted by tinyone
Cheers,

Derivative f'z(x,y,z) = 2(z+3)^4 + ln(xyz)

So keeping y and x constant would give me: f'z= 8(z+3)^3 + (1/xyz) * xy
Of course, you would want to cancel the "xy" terms in numerator and denominator of that last fraction to get
$f_z= 8(z+ 3)^3+ \frac{1}{z}$
You could see that more easily by writing f(x, y, z()= 2(z+ 3)^4+ ln(xyz)= 2(z+ 3)^4+ ln(x)+ ln(y)+ ln(z).

Oh, and don't write "f'z" for both the function and its partial derivative!

In fact, it is not a good idea to use the ' with partial derivatives. You problem is to find the derivative, with respect to z,
of f(x,y,z)= 2(z+ 3)^4+ ln(xyz) and the correct answer, in simplest form, is f_z= 8(z+ 3)^3+ 1/z.