# How do you find the derivative of y=5/(x-3)^2?

Jun 3, 2018

$y ' = \frac{- 10}{x - 3} ^ 3$

#### Explanation:

Use Quotient Rule :

$\textcolor{red}{y ' = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2}$

$y = \frac{5}{x - 3} ^ 2$

$y ' = \frac{{\left(x - 3\right)}^{2} \cdot 0 - 5 \cdot 2 \left(x - 3\right) \cdot 1}{x - 3} ^ 4$

$y ' = \frac{- 10 \left(x - 3\right)}{x - 3} ^ 4$

$y ' = \frac{- 10}{x - 3} ^ 3$

Jun 3, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{10}{x - 3} ^ 3$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$y = \frac{5}{x - 3} ^ 2 = 5 {\left(x - 3\right)}^{-} 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 10 {\left(x - 3\right)}^{-} 3 \times \frac{d}{\mathrm{dx}} \left(x - 3\right)$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{10}{x - 3} ^ 3$