How do you find the derivative of the function: $\sqrt{1 - 9 x^{2}} arccos (3 x)$ $sqrt(1-9x^2)arccos(3x)$ ?

Question

4229 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-21T02:42:22

$y = \sqrt{1 - 9 x^{2}} arccos (3 x)$ $y=sqrt(1-9x^2)arccos(3x)$

Let $3 x = cos θ$ $" "3x=costheta$

So Differentiating w r to x we have $3 = - sin θ \frac{d θ}{d x}$ $" " 3=-sintheta(d theta)/(dx)$

$\Rightarrow \frac{d θ}{d x} = - \frac{3}{sin θ}$ $=>(d theta)/(dx)=-3/sintheta$

Now

$y = \sqrt{1 - 9 x^{2}} arccos (3 x)$ $y=sqrt(1-9x^2)arccos(3x)$

$\Rightarrow y = \sqrt{1 - {cos}^{2} θ} arccos (cos θ)$ $=>y=sqrt(1-cos^2theta)arccos(costheta)$

$\Rightarrow y = θ sin θ$ $=>y=thetasintheta$

Differentiating w r to $θ$ $theta$ we have

$\frac{d y}{d θ} = θ cos θ + sin θ$ $(dy)/(d theta)=thetacostheta+sintheta$

So

$\frac{d y}{d x} = \frac{d y}{d θ} \times \frac{d θ}{d x}$ $(dy)/(dx)=(dy)/(d theta)xx(d theta)/(dx)$

$\Rightarrow \frac{d y}{d x} = (θ cos θ + sin θ) \times (- \frac{3}{sin θ})$ $=>(dy)/(dx)=(thetacostheta+sintheta)xx(-3/sintheta)$

$\Rightarrow \frac{d y}{d x} = (- 3 θ \frac{cos θ}{sin θ} - 3)$ $=>(dy)/(dx)=(-3thetacostheta/sintheta-3)$

$\Rightarrow \frac{d y}{d x} = - 3 (\frac{θ cos θ}{\sqrt{1 - {cos}^{2} θ}} + 1)$ $=>(dy)/(dx)=-3((thetacostheta)/sqrt(1-cos^2theta)+1)$

$\Rightarrow \frac{d y}{d x} = - 3 (\frac{3 x {cos}^{- 1} (3 x)}{\sqrt{1 - 9 x^{2}}} + 1)$ $=>(dy)/(dx)=-3((3xcos^-1(3x))/sqrt(1-9x^2)+1)$

How do you find the derivative of the function: sqrt(1-9x^2)arccos(3x)√1−9x2arccos(3x)sqrt(1-9x^2)arccos(3x)?