# How do you differentiate f(x)=sqrt(1-(3x-3)^2 using the chain rule.?

Oct 15, 2017

The derivative is $\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{9 \left(x - 1\right)}{\sqrt{1 - {\left(3 x - 3\right)}^{2}}}}$

#### Explanation:

Let the function given be $y$ such that,

$y = \sqrt{1 - {\left(3 x - 3\right)}^{2}}$

Now, it is better to simplify the term $\left(1 - {\left(3 x - 3\right)}^{2}\right)$.

$\therefore y = \sqrt{1 - 9 {\left(x - 1\right)}^{2}}$

:.y=sqrt(1-9(x^2-2x+1)

$\therefore y = \sqrt{1 - 9 {x}^{2} + 18 x - 9}$

$\therefore y = \sqrt{- 9 {x}^{2} + 18 x - 8}$

$\therefore y = {\left(- 9 {x}^{2} + 18 x - 8\right)}^{\frac{1}{2}}$

Now differentiating w.r.t $x$ we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \cdot {\left(- 9 {x}^{2} + 18 x - 8\right)}^{\frac{1}{2} - 1} \times \frac{d}{\mathrm{dx}} \left(- 9 {x}^{2} + 18 x - 8\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{- 9 {x}^{2} + 18 x - 8}} \times \left(- 18 x + 18 - 0\right)$

Now, simplifying the equation $\rightarrow$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{18}{2 \sqrt{1 - {\left(3 x - 3\right)}^{2}}} \times \left(x - 1\right)$

$\therefore \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{9 \left(x - 1\right)}{\sqrt{1 - {\left(3 x - 3\right)}^{2}}}}$. (Answer)

Hope it Helps:)