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By definition of the derivative,
[tex]\displaystyle t'(x) = \lim_{h\to0} \frac{t(x+h) - t(x)}h[/tex]
[tex]\displaystyle t'(x) = \lim_{h\to0} \frac{\sqrt{-3(x+h)-7} - \sqrt{-3x-7}}h[/tex]
Rationalize the numerator by multiplying the fraction uniformly by its conjugate:
[tex]\displaystyle t'(x) = \lim_{h\to0} \frac{\sqrt{-3(x+h)-7} - \sqrt{-3x-7}}h \times \dfrac{\sqrt{-3(x+h)-7} + \sqrt{-3x - 7}}{\sqrt{-3(x+h)-7} + \sqrt{-3x - 7}}[/tex]
[tex]\displaystyle t'(x) = \lim_{h\to0} \frac{\left(\sqrt{-3(x+h)-7}\right)^2 - \left(\sqrt{-3x-7}\right)^2}{h \left(\sqrt{-3(x+h)-7} + \sqrt{-3x - 7}\right)}[/tex]
[tex]\displaystyle t'(x) = \lim_{h\to0} \frac{(-3(x+h)-7) - (-3x-7)}{h \left(\sqrt{-3(x+h)-7} + \sqrt{-3x - 7}\right)}[/tex]
[tex]\displaystyle t'(x) = \lim_{h\to0} \frac{-3h}{h \left(\sqrt{-3(x+h)-7} + \sqrt{-3x - 7}\right)}[/tex]
[tex]\displaystyle t'(x) = -3 \lim_{h\to0} \frac1{\sqrt{-3(x+h)-7} + \sqrt{-3x - 7}}[/tex]
The remaining limand is continuous at h = 0, so we can substitute h = 0 directly and get a limit/derivative of
[tex]\displaystyle t'(x) = -\frac3{\sqrt{-3(x+0)-7} + \sqrt{-3x - 7}} = \boxed{-\frac3{2\sqrt{-3x-7}}}[/tex]