Can anyone help me to find the derivative of this function? I know I have to use the quotient rule: $\dfrac{f(x)}{g(x)}=\dfrac{f′(x)g(x)−f(x)g′(x)}{(g(x))^2}$, but I don't know how I use this when the function is:
$$f(x,y) = \frac{7y + x^2}{1+y^2}$$
$f_x (x,y) = ?$
$f_y (x,y) = ?$
On
To find the partial derivatives, treat the variable your differentiating with respect to as your variable and treat all the other variables as constants.
So, when we are finding the partial derivative with respect with x, we differentiate with respect with x and treat y (the other variable) as a constant. Same principle goes with differentiating with respect with y.
So our function is $$f(x,y) = \frac{7y+x^2}{1+y^2} = \frac{7y}{1+ y^2} + \frac{x^2}{1+ y^2}$$
So $$f_x = 0 + \frac{2x}{1+y^2} = \frac{2x}{1+y^2}$$
Continuing on: $$f_y = \frac{7(1+y^2)-2y(7y+x^2)}{(1+y^2)^2} = \frac{7-7y^2+2x^2y}{(1+y^2)^2}$$
While taking partial derivatives you should think of the given function as a function with one parameter depending on a single variable, in other words:
$$f_x(y):=f(x,y)=:f_y(x)$$
Where once you regard $x$ as a parameter and the other you do the same for $y$
[It might make this more understandable or not, it's up to you to decide]
Now that we have functions in just one variable you can apply the definition of derivative for functions in one variable, so:
$$\frac{\partial f(x,y)}{\partial x}=\frac{df_y(x)}{dx}$$
And analogously
$$\frac{\partial f(x,y)}{\partial y}=\frac{df_x(y)}{dy}$$
Now for you specific example:
$$f(x,y)= \frac{(7y + x^2)}{(1+y^2)}$$
So
$$\frac{\partial f(x,y)}{\partial x}=\frac{2x}{1+y^2}$$
(because the derivative of the "parameter " $y$ with respect to $x$ is zero)
Can you manage to do the partial derivative wrt to y?