Differential Calculus> y= cos^-1(2x + 3 sqrt(1-x^2) / sqrt(13)) ...

y=cos^-1(2x + 3 sqrt(1-x^2) / sqrt(13)) find dy/dx

taniska , 10 Years ago

Grade 12

2 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

Hello Student,
Please find answer to your question

$y = cos^{-1}(\frac{2x+3\sqrt{1-x^{2}}}{\sqrt{13}})$
Apply the chain rule
$\frac{dy}{dx} = \frac{d(cos^{-1}u)}{dx}.\frac{du}{dx}$
$u = \frac{2x+3\sqrt{1-x^{2}}}{\sqrt{13}}$
$u^{2} = \frac{4x^2+9(1-x^{2})+12x\sqrt{1-x^{2}}}{13}$
$u^{2} = \frac{9-5x^{2}+12x\sqrt{1-x^{2}}}{13}$
$1-u^{2} = 1-\frac{9-5x^{2}+12x\sqrt{1-x^{2}}}{13}$
$1-u^{2} = \frac{4+5x^{2}-12x\sqrt{1-x^{2}}}{13}$

$\frac{du}{dx} = \frac{2+3.\frac{-x}{\sqrt{1-x^{2}}}}{\sqrt{13}}$
$\frac{du}{dx} = \frac{2\sqrt{1-x^{2}}-3x}{\sqrt{13(1-x^{2})}}$
$\frac{dy}{dx} = \frac{d(cos^{-1}u)}{dx}.\frac{du}{dx}$
$\frac{dy}{dx} = \frac{-1}{\sqrt{1-u^{2}}}.\frac{du}{dx}$
$\frac{dy}{dx} = \frac{-\sqrt{13}}{\sqrt{4+5x^{2}-12x\sqrt{1-x^{2}}}}.\frac{du}{dx}$
$\frac{dy}{dx} = \frac{-\sqrt{13}}{\sqrt{4+5x^{2}-12x\sqrt{1-x^{2}}}}.\frac{2\sqrt{1-x^{2}}-3x}{\sqrt{13(1-x^{2})}}$
$\frac{dy}{dx} = \frac{3x-2\sqrt{1-x^{2}}}{\sqrt{4+5x^{2}-12x\sqrt{1-x^{2}}}}.\frac{1}{\sqrt{1-x^{2}}}$
$\frac{dy}{dx} = \frac{3x-2\sqrt{1-x^{2}}}{\sqrt{(3x-2\sqrt{1-x^{2}})^{2}}}.\frac{1}{\sqrt{1-x^{2}}}$ $\frac{dy}{dx} = \frac{3x-2\sqrt{1-x^{2}}}{|(3x-2\sqrt{1-x^{2}})|}.\frac{1}{\sqrt{1-x^{2}}}$
$\frac{dy}{dx} = \frac{\pm 1}{\sqrt{1-x^{2}}}$