I need help finding the critical values of this function. So $h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$. So I need to set $h'(t)=0$. So for $h'(t)$ the fattest I've gotten to simplifying os $h'(t)=\frac{3}{4 \sqrt[4]{t}}-\frac{7}{4\sqrt[4]{t^3}}$ and that is as farthest as I can simplify. So i'm having a had time having $h'=0$ So could someone show me how to properly simplify and find the critical values step by step I would immensely appreciate it. In the most recent material we have covered in class ir seems that my biggest struggle comes from simplifying completely. Thanks in advance for the help. 

Hint
To make your life easier, just define $x=t^{\frac{1}{4}}$. So $h(t)=x^3-7x$ and then $$\frac{dh}{dt}=\frac{dh}{dx}\frac{dx}{dt}=(3x^2-7)\frac{dx}{dt}$$ So,$h'(t)=0$ if $x^2=\frac{7}{3}$ that is to say $\sqrt t=\frac{7}{3}$ then $t=\frac{49}{9}$.
