# Finding derivatives , where am I going wrong?

What I've done:

$$\frac{df}{dx} = 15y^3 \cos \left(3x\right)- 10xy^6 +2e^x \sin \left(y\right)$$

$$\frac{d^2f}{dx^2} = -45 y^3 \sin \left(3x\right) - 10y^6 + 2e^x \sin \left(y\right)$$

$$\frac{df}{dy} = 15 y^3 \cos \left(3x\right) - 30 x^2 y^5 +2 e^x \cos \left(y\right)$$

$$\frac{df^2}{dy^2} = 45y^2 \cos \left(3x\right) -150 x^2 y^4 -2e^x \sin \left(y\right)$$

When I go to do the last part they are not equaling

When I derive $$\frac{df^2}{dy^2}$$ in terms of $$x$$ I get $$-135y^2 \sin \left(3x\right) - 300x y^4 - 2e^x \sin \left(y\right)$$

When I derive $$\frac{df^2}{dx^2}$$ in terms of $$y$$ I get

$$-135 y^2 \sin \left(3x\right) - 60y^5 - 2e^x \cos \left(y\right)$$

What am I doing wrong here? When I derive the first equations they equal each other.

• You made a careless error in $\partial f/\partial y$. – David Mar 7 '16 at 22:45

You made a little mistake in the derivative for $$\frac{df}{dy}$$ : it should be $$5⋅3y^2⋅\sin \left(3x\right)-5 ⋅ x^2⋅ 6⋅ y^5 - 2 e^x \cos \left(y\right) = 15 y^2 \sin \left(3x\right)-30 x^2y^5-2e^x\cos \left(y\right)$$
• Well, I'm still seeing errors in your question : you still derivate $\sin(3x)$ in your $\frac{d}{dy}$ and $\frac{d^2}{dy^2}$ derivative, but $\sin(3x)$ doesn't depend on $y$ so it's just a constant when you are considering $dy$. Please edit your question and add an "EDIT" section at the end with your new calculations. – Lery Mar 10 '16 at 9:46
You should get that $$\frac {\partial^2 f}{\partial x\partial y} =\frac {\partial^2 f}{\partial y\partial x}$$
There is no reason to expect $$\frac {\partial}{\partial y}\frac {\partial^2 f}{\partial^2 x}$$ should equal $$\frac {\partial}{\partial x}\frac {\partial^2 f}{\partial^2 y}$$
$$\frac {\partial}{\partial y}(15y^3\cos 3x - 10xy^6 + 2e^x\sin y) = 45y^2\cos 3x - 60 xy^5 + 2e^x\cos y$$
$$\frac {\partial}{\partial x}(15y^2\sin 3x - 30x^2y^5 + 2e^x\cos y) = 45y^2\cos 3x - 60 xy^5 + 2e^x\cos y$$