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

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Answer 1 · 2017-10-15T15:13:09

The derivative is $\frac{d y}{d x} = - \frac{9 (x - 1)}{\sqrt{1 - {(3 x - 3)}^{2}}}$ $color(red)(dy/dx=-(9(x-1))/sqrt(1-(3x-3)^2))$

Let the function given be $y$ $y$ such that,

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

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

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

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

$:.y=sqrt(1-9x^2+18x-9)$

$:.y=sqrt(-9x^2+18x-8)$

$:.y=(-9x^2+18x-8)^(1/2)$

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

$dy/dx=1/2cdot(-9x^2+18x-8)^(1/2-1)xxd/dx(-9x^2+18x-8)$

$:.dy/dx=1/(2sqrt(-9x^2+18x-8))xx(-18x+18-0)$

Now, simplifying the equation $rarr$

$:.dy/dx=-18/(2sqrt(1-(3x-3)^2))xx(x-1)$

$:.color(red)(dy/dx=-(9(x-1))/sqrt(1-(3x-3)^2))$ . (Answer)

Hope it Helps:)

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