Fisher's equation solution step

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There is a step in finding steady states for fisher's equation that I cannot follow.

The equation is given as

$\delta_tu=D\delta_{xx}u+ru\Big(1-\frac{u}{K}\Big)$, where $D,r,K>0$.

The first step is to plug $u=\varphi(x-ct)$ into this equation.

The result is $-c\varphi'=D\varphi''+r\varphi\Big(1-\frac{\varphi}{K}\Big)$.

This step I do not understand. For example, putting $\varphi(x-ct)$ into $\delta_tu$ gives me $-c\varphi$ instead of $-c\varphi'$? $\big(\frac{d}{dt}\varphi(x-ct)=-c\varphi\big)$

Thank you for any help