# How do you find the derivative for f(t)= sqrt(t) / (-4t-6)?

Jul 31, 2015

${y}^{'} = - \frac{1}{4} \cdot \frac{3 - 2 t}{\sqrt{t} \cdot {\left(2 t + 3\right)}^{2}}$

#### Explanation:

You can differentiate this function by using the quotient rule, which tells you that the derivative of afunction that takes the form

$y = f \frac{x}{g} \left(x\right)$

can be calculated by using

color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2, with $g \left(x\right) \ne 0$.

Your original function looks like this

$f \left(t\right) = \frac{\sqrt{t}}{- 4 t - 6}$

You can rewrite this function as

$f \left(t\right) = \frac{1}{- 2} \cdot \frac{\sqrt{t}}{2 t + 3} = - \frac{1}{2} \cdot \frac{\sqrt{t}}{2 t + 3}$

The derivative of $f \left(t\right)$ will thus be

$\frac{d}{\mathrm{dt}} \left(f\right) = - \frac{1}{2} \cdot \left[\frac{\left[\frac{d}{\mathrm{dt}} \left(\sqrt{t}\right)\right] \cdot \left(2 t + 3\right) - \sqrt{t} \cdot \frac{d}{\mathrm{dx}} \left(2 t + 3\right)}{2 t + 3} ^ 2\right]$

${f}^{'} = - \frac{1}{2} \cdot \frac{\frac{1}{2 \sqrt{t}} \cdot \left(2 t + 3\right) - 2 \sqrt{t}}{2 t + 3} ^ 2$

${f}^{'} = - \frac{1}{2 {\left(2 t + 3\right)}^{2}} \cdot \left(\frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \sqrt{t}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} t + \frac{3}{2 \sqrt{t}} - 2 \sqrt{t}\right)$

${f}^{'} = - \frac{1}{2 {\left(2 t + 3\right)}^{2}} \cdot \left(\sqrt{t} + \frac{3}{2 \sqrt{t}} - 2 \sqrt{t}\right)$

${f}^{'} = - \frac{1}{2 {\left(2 t + 3\right)}^{2}} \cdot \frac{3 - \sqrt{t} \cdot 2 \sqrt{t}}{2 \sqrt{t}}$

${f}^{'} = \textcolor{g r e e n}{- \frac{1}{4} \cdot \frac{3 - 2 t}{\sqrt{t} \cdot {\left(2 t + 3\right)}^{2}}}$