# Find all the directional derivative of the function $f(x, y)=y^{2} e^{4 x}$

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?

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

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 $$(4,-2)$$.