Find $f_y$ of $e^{x^2 + 2y^2 + 3z^2}$

How do I do this ?

I am trying to follow the formula of -

$\frac{d}{dx} e^{ax +b} = ae^{ax+b} $

Since I am differentiating the independent variable of y,

Why isn’t the partial derivative $ (f_y)$ of $e^{x^2 + 2y^2 + 3z^2} = 2e^{x^2 + 2y^2 + 3z^2} \frac{\partial}{\partial y} (x^2 + 2y^2 + 3z^2) $ ?

In fact the workings is -

$e^{x^2 + 2y^2 + 3z^2} \frac{\partial}{\partial y} (x^2 + 2y^2 + 3z^2) $

Why isn’t the ‘2’ infront if $e$ ?


3 Answers 3


Note that by the chain rule, $$\frac{d}{dy} \left(e^{ag(y) +b}\right) = ae^{ag(y)+b}\cdot g'(y).$$ Hence the partial derivative of $f(x,y,z)=e^{x^2 + 2y^2 + 3z^2}=e^{ag(y) +b}$ with respect to $y$ is $$f_y(x,y,z)=2e^{x^2 + 2y^2 + 3z^2} \frac{\partial}{\partial y} (y^2).$$ where we applied the above rule with $a=2$, $g(y)=y^2$ and $b=x^2+3z^2$.

  • $\begingroup$ oh.. so in my case , $ f(x) = x^2 + 2y^2 + 3z^2 $ $\endgroup$
    – user185692
    Aug 12, 2018 at 9:25
  • $\begingroup$ @user185692 I added a few details. Is it clear now? $\endgroup$
    – Robert Z
    Aug 12, 2018 at 9:27
  • $\begingroup$ yup! Thanks!! Managed to see where went wrong $\endgroup$
    – user185692
    Aug 12, 2018 at 9:29

You don't have a function of the form $f(x)=e^{ax+b}$; but rather one of the form $f(x)=e^{ax^2+b}$. (It is written in terms of $y$ rather than $x$, but of course this doesn't matter).

Now (for the latter $f$), $f_x(x)=e^{ax^2+b}\cdot \frac d{dx}(ax^2+b)=e^{ax^2+b}\cdot 2ax$, where I have used that the derivative of $e^x$ is $e^x$, and the chain rule...

Make the appropriate substitutions...


You need to apply the chain rule the following way: Let $e^{x^2+2y^2+3z^2} = f(g(x,y,z))$ where $f(a) = e^a$ and $g(x,y,z) = x^2+2y^2+3z^2$. $\frac {\partial f \circ g} {\partial y}(x,y,z) = (\frac {\partial} {\partial y} f(g(x,y,z))) \cdot (\frac {\partial} {\partial y}g(x,y,z)) = (e^{x^2+2y^2+3z^2}) \cdot (4y) = 4y \cdot e^{x^2+2y^2+3z^2}$.


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