Grade 12Differential Calculus

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

taniska

11 Years agoGrade 12

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

$y = cos^{-1}(\frac{\sqrt{1+x}+\sqrt{1-x}}{2})$
Apply chain rule, we have
$\frac{dy}{dx} = \frac{dcos^{-1}u}{dx}.\frac{du}{dx}$
$u = \frac{\sqrt{1+x}+\sqrt{1-x}}{2}$
$u^{2} = \frac{2+2\sqrt{1-x^{2}}}{4}$
$u^{2} =\frac{1+\sqrt{1-x^{2}}}{2}$

$\frac{dy}{dx} = \frac{-1}{\sqrt{1-u^{2}}}.\frac{du}{dx}$
$\frac{dy}{dx} = \frac{-\sqrt{2}}{\sqrt{1-\sqrt{1-x^{2}}}}.\frac{1}{2}.(\frac{1}{2\sqrt{x+1}}+\frac{-1}{2\sqrt{1-x}})$
$\frac{dy}{dx} = \frac{-\sqrt{2}}{\sqrt{1-\sqrt{1-x^{2}}}}.\frac{1}{4}.(\frac{1}{\sqrt{x+1}}+\frac{-1}{\sqrt{1-x}})$ $\frac{dy}{dx} = \frac{-\sqrt{2}}{\sqrt{1-\sqrt{1-x^{2}}}}.\frac{1}{4}.(\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x^{2}}})$