Calculate the third partial derivative If $u = vt^{\frac{1}{3}}$ and $v=h(y), y=it^{\frac{1}{3}}\sin(iwx)$ I need to determine $ u_{xxx}$ and $u_t$
I have :
\begin{align}
u_x &amp;= u_v \cdot v_y \cdot y_x\\
&amp;= t^{\frac{1}{3}}v_y(-wt^{\frac{1}{3}}\cos(iwx))\\
&amp;=-wt^{\frac{2}{3}}\cos(iwx)v_y
\end{align}
You have $y=it^{1/2}\sin iwx$. So $\frac{dy}{dx}=-wt^{1/2}\cos iwx$. Then
$$\begin{split}u_x&amp;=-wt^{2/3}h'(y)\cos iwx\end{split}$$
To get from $u_x$ to $u_{xx}$ you need to use the product rule. $u_{xx}=-wt^{2/3}\left[\left(\frac{d}{dx}h'(y)\right)\cos iwx+h'(y)\frac{d}{dx}\cos iwx\right]$. Now you can apply the chain rule now that you've applied the product rule.
$$\begin{split}u_{xx}&amp;=-wt^{2/3}\left[h''(y)(-wt^{1/2}\cos^2 iwx)+h'(y)(-iw\sin iwx)\right]\\
&amp;=w^2t^{2/3}\left[h''(y)t^{1/2}\cos^2 iwx+h'(y)i\sin iwx\right]\\
u_{xxx}&amp;=w^2t^{2/3}\left[h'''(y)t^{1/2}(-wt^{1/2}\cos^3 iwx)+h''(y)t^{1/2}2\cos iwx (-iw\sin iwx)+h''(y)i\sin iwx (-wt^{1/2}\cos iwx)+h'(y)i^2w\cos iwx\right]\\
&amp;=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-2iwt^{1/2}h''(y)\cos iwx \sin iwx-it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&amp;=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-3it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&amp;=-w^3t^{2/3}\cos iwx\left[t^{2/3}h'''(y)\cos^2 iwx+3it^{1/2}h''(y)\sin iwx +h'(y)\right]\end{split}$$
You have $y=it^{1/2}\sin iwx$. So $\frac{dy}{dx}=-wt^{1/2}\cos iwx$. Then
$$\begin{split}u_x&amp;=-wt^{2/3}h'(y)\cos iwx\end{split}$$
To get from $u_x$ to $u_{xx}$ you need to use the product rule. $u_{xx}=-wt^{2/3}\left[\left(\frac{d}{dx}h'(y)\right)\cos iwx+h'(y)\frac{d}{dx}\cos iwx\right]$. Now you can apply the chain rule now that you've applied the product rule.
$$\begin{split}u_{xx}&amp;=-wt^{2/3}\left[h''(y)(-wt^{1/2}\cos^2 iwx)+h'(y)(-iw\sin iwx)\right]\\
&amp;=w^2t^{2/3}\left[h''(y)t^{1/2}\cos^2 iwx+h'(y)i\sin iwx\right]\\
u_{xxx}&amp;=w^2t^{2/3}\left[h'''(y)t^{1/2}(-wt^{1/2}\cos^3 iwx)+h''(y)t^{1/2}2\cos iwx (-iw\sin iwx)+h''(y)i\sin iwx (-wt^{1/2}\cos iwx)+h'(y)i^2w\cos iwx\right]\\
&amp;=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-2iwt^{1/2}h''(y)\cos iwx \sin iwx-it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&amp;=w^2t^{2/3}\left[-wt^{2/3}h'''(y)\cos^3 iwx-3it^{1/2}wh''(y)\sin iwx \cos iwx-h'(y)w\cos iwx\right]\\
&amp;=-w^3t^{2/3}\cos iwx\left[t^{2/3}h'''(y)\cos^2 iwx+3it^{1/2}h''(y)\sin iwx +h'(y)\right]\end{split}$$