What is the derivative of  tan^-1 ((1.5)/x) - tan^-1 (0.5)?

Feb 25, 2018

$- \frac{6}{4 {x}^{2} + 9}$

Explanation:

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(\frac{1.5}{x}\right) - {\tan}^{-} 1 \left(0.5\right)\right) = \frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(1.5 {x}^{-} 1\right) - {\tan}^{-} 1 \left(0.5\right)\right)$

When differentiating, $- {\tan}^{-} 1 \left(0.5\right)$ goes away as it's just a constant value. The derivative of any constant is $0.$

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(1.5 {x}^{-} 1\right) - {\tan}^{-} 1 \left(0.5\right)\right) = \left(\frac{1}{1 + {\left(1.5 {x}^{-} 1\right)}^{2}}\right) \cdot \frac{d}{\mathrm{dx}} \left(1.5 {x}^{-} 1\right)$

Using the Chain Rule along with the fact that

$\frac{d}{\mathrm{dx}} {\tan}^{-} 1 \left(x\right) = \frac{1}{1 + {x}^{2}}$. Here, we have $1.5 {x}^{-} 1$ as the argument of the arctangent, so instead of ${x}^{2}$, we will have ${\left(1.5 {x}^{-} 1\right)}^{2}$ in the denominator.

$\left(\frac{1}{1 + {\left(1.5 {x}^{-} 1\right)}^{2}}\right) \cdot \frac{d}{\mathrm{dx}} \left(1.5 {x}^{-} 1\right) = \frac{1}{1 + 2.25 {x}^{-} 2} \cdot - 1.5 {x}^{-} 2$

Simplify:

$- \frac{1.5}{{x}^{2} \left(1 + \frac{2.25}{x} ^ 2\right)} = - \frac{1.5}{{x}^{2} + \frac{2.25 {\cancel{x}}^{2}}{\cancel{x}} ^ 2} = - \frac{1.5}{{x}^{2} + 2.25}$

$1.5 = \frac{3}{2}$
$2.25 = \frac{9}{4}$

Getting rid of the decimals:

$- \frac{1.5}{{x}^{2} + 2.25} = - \frac{\frac{3}{2}}{{x}^{2} + \frac{9}{4}} = - \frac{3}{2 \left({x}^{2} + \frac{9}{4}\right)} = - \frac{3}{2 {x}^{2} + \frac{9}{2}} = - \frac{3}{\frac{4 {x}^{2} + 9}{2}} = - 3 \cdot \left(\frac{2}{4 {x}^{2} + 9}\right) = - \frac{6}{4 {x}^{2} + 9}$