# How do you find the derivative for (t^1.7 + 8)/( t^1.4 + 6)?

Jul 30, 2015

You can use the quotient rule.

#### Explanation:

You can differentiate a function $f \left(x\right)$ that can be written as the quotient of two other functions, $g \left(x\right)$ and $h \left(x\right)$, by using the quotient rule

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

$g \left(t\right) = {t}^{1.7} + 8$

and

$h \left(y\right) = {t}^{1.4} + 6$

This means that the derivative of your function $f \left(t\right) = g \frac{t}{h \left(t\right)}$ will be

${f}^{'} \left(t\right) = \frac{\frac{d}{\mathrm{dt}} \left({t}^{1.7} + 8\right) \cdot \left({t}^{1.4} + 6\right) - \left({t}^{1.7} + 8\right) \cdot \frac{d}{\mathrm{dt}} \left({t}^{1.4} + 6\right)}{{t}^{1.4} + 6} ^ 2$

${f}^{'} \left(t\right) = \frac{1.7 \cdot {t}^{0.7} \cdot \left({t}^{1.4} + 6\right) - \left({t}^{1.7} + 8\right) \cdot 1.4 \cdot {t}^{0.4}}{{t}^{1.4} + 6} ^ 2$

This is equivalent to

${f}^{'} \left(t\right) = \frac{\frac{1}{10} \left(17 \cdot {t}^{2.1} + 102 \cdot {t}^{0.7} - 14 \cdot {t}^{2.1} - 112 \cdot {t}^{0.4}\right)}{{t}^{1.4} + 6} ^ 2$

${f}^{'} \left(t\right) = \frac{3 \cdot {t}^{2.1} + 102 \cdot {t}^{0.7} - 112 \cdot {t}^{0.4}}{10 {\left({t}^{1.4} + 6\right)}^{2}}$

Finally, you can write

f^'(t) = color(green)(1/10 * (t^0.4(3 * t^1.7 + 102 * t^0.3 - 112))/(t^1.4 + 6)^2