# Directional derivative of $z(x,y)=yg\left(\frac{x}y\right)+\frac{8y^3}x$ on the unit circle

Hi can someone help me with this question?

Let $g(t)$ be differentiable for all $t$ with $g\left(\frac12\right)=1$ and $$z(x,y)=yg\left(\frac{x}y\right)+\frac{8y^3}x$$ $(x,y)$ is a point on the circle $x^2+y^2=1$ and $x\ne0$. What is the directional derivative of the function $z(x,y)$ in the point $(x,y)$, in the direction of the vector that points to the center of the circle? (the answer should be a function of $x,y,z$ and not $g$).

## 2 Answers

Since $\;g\;$ is differentiable everywhere so is $\;z\;$ at any point $\;xy\neq0\;$ , and thus on the unit circle, and we can evaluate the directional derivative by means of the gradient:

$$\nabla z=\left(\,g'\left(\frac xy\right)-\frac{8y^3}{x^2}\,,\,\,g\left(\frac xy\right)-\frac xyg'\left(\frac xy\right)+\frac{24y^2}{x}\,\right)$$

and then the wanted directional derivative at a point $\;(x,y)\;$ on the unit circle and in the direction of $\;(-x,-y)\;$ (the vector from $\;(x,y)\;$ towards the circle's center = the origin) is:

$$\nabla f\bullet (-x,-y)=\color{red}{-xg'\left(\frac xy\right)}+\frac{8y^3}x-yg\left(\frac xy\right)+\color{red}{xg'\left(\frac xy\right)}-\frac{24y^3}{x}=$$

$$=-yg\left(\frac xy\right)+\frac{16y^3}x=-z-\frac{8y^3}x$$

The directional derivative of $z$ at a point $(x,y)$ in the direction of a vector $v$ is defined at the dot product $\cdot$ of $\nabla z$ and $v$, where $\nabla z$ denotes the vector with the partial derivatives of $z$. Hence, \begin{align}\frac{\partial }{\partial x}z(x,y)&=g'\left(\frac{x}y\right)-\frac{8y^3}{x^2}\\[0.2cm]\frac{\partial }{\partial y}z(x,y)&=g\left(\frac{x}y\right)-\frac{x}{y}g'\left(\frac{x}{y}\right)+\frac{24y^2}{x}\end{align} The center of the circle is $(0,0)$, so at point $(x,y)$ the vector of length $1$ that points to the center of the circle is simply $-(x,y)$ (this is because the circle is of radius $1$), so \begin{align}\left(\frac{\partial }{\partial x}z(x,y),\frac{\partial }{\partial y}z(x,y)\right)\cdot(-x,-y)&=-xg'\left(\frac{x}y\right)+\frac{8y^3}{x}-yg\left(\frac{x}y\right)+xg'\left(\frac{x}{y}\right)-\frac{24y^3}{x}\\[0.2cm]&=-yg\left(\frac{x}y\right)-\frac{16y^3}{x}=-z(x,y)-\frac{8y^3}{x}\end{align} but I did not use anywhere that $g(1/2)=1$.