Answer:
h'(t) = (√t)(1 -5t²)/(2t)
Explanation:
First of all, your product rule needs to be written correctly.
(f(x)g(x))' = f'(x)g(x) +f(x)g'(x)
You have ...
[tex]h(t)=\sqrt{t}(1-t^2)\\\\f(t)=\sqrt{t} \qquad f'(t)=\dfrac{1}{2\sqrt{t}}=\dfrac{\sqrt{t}}{2t}\\\\g(t)=1-t^2 \qquad g'(t)=-2t\\\\h'(t)=f'(t)g(t)+f(t)g'(t)=\dfrac{\sqrt{t}}{2t}(1-t^2)+\sqrt{t}(-2t)\\\\h'(t)=\dfrac{\sqrt{t}(1-t^2-4t^2)}{2t}=\dfrac{\sqrt{t}(1-5t^2)}{2t}[/tex]
