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I would like to find all the directional derivative of the function $f(x, y)=y^{2} e^{4 x}$ in $P=(0,-1)$.

I know that the definition of directional derivative, fixing a direction $\vec{u}=(a,b)$, is:

$$D_{\vec{u}} f\left(x_{0}, y_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h a, y_{0}+h b\right)-f\left(x_{0}, y_{0}\right)}{h}$$

Equally, it can be expressed as follows:

$$D_{\vec{u}} f(x, y)=\nabla f(x,y)^T\vec{u}=\frac{\partial f}{\partial x} a+\frac{\partial f}{\partial y} b=\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \cdot(a, b)$$

Now, I've calculated the partial derivatives of $f(x,y)$ which are: $f_x=4y^2e^{4x}, f_y=2ye^{4x}$.

Now, my textbook's result is: $D_{\vec{u}}f(P)=(cos\theta, sin\theta)\cdot (4, -2)$. How does it get there?

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  • $\begingroup$ Your second formulation of a directional derivative is only valid when $f$ is differentiable at $(a,b)$, which, fortunately, is the case here. $\endgroup$
    – amd
    Jan 29, 2020 at 19:03

1 Answer 1

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The vector $\vec{u}$ is a unit vector. $\vec{u}=(\cos \theta, \sin \theta).$

If you evaluate $(f_x,f_y)$ at $P$, you get $(4,-2)$.

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