Question

If y = sin-1 x + cos-1x, find a
dy/dx​

Answer 1

Step-by-step explanation:

hope it helps

0is the answer

Answer 2

$\dfrac{dy}{dx}=0$

Step-by-step explanation:

We have,

$y = sin^{-1} x + cos^{-1} x$           ........(1)

To find, $\dfrac{dy}{dx}=?$

Differentiating (1) w.r.t. x, we get

$\dfrac{dy}{dx}= \dfrac{d(sin^{-1} x)}{dx}  +\dfrac{d(cos^{-1} x)}{dx}$

⇒ $\dfrac{dy}{dx}= \dfrac{1}{\sqrt{1-x^{2}}} +\dfrac{-1}{\sqrt{1-x^{2}}}$

[ ∵ $sin^{-1} x = \dfrac{1}{\sqrt{1-x^{2}}}$ and $\cos ^{-1}x  = \dfrac{-1}{\sqrt{1-x^{2}}}$]

⇒ $\dfrac{dy}{dx}= \dfrac{1}{\sqrt{1-x^{2}}} -\dfrac{1}{\sqrt{1-x^{2}}}$

⇒ $\dfrac{dy}{dx}= 0</strong></p><p><strong>Hence, [tex]\dfrac{dy}{dx}=0$