Question

find dy/dx:. y=sin^-1(x^2+2x)​

Answer 1

Step-by-step explanation:

Given that

$y = { \sin( {x}^{2} + 2x) }^{ - 1}$

Differentiate with respect to x

$\frac{dy}{dx} = \frac{1}{ \sqrt{1 - { ({x}^{2} + 2x)}^{2} } } \times \frac{d}{dx} ( {x}^{2} + 2x) \\ \: =\frac{1}{ \sqrt{1 - { ({x}^{2} + 2x)}^{2} } } \times (2x + 2) \\ \: =\frac{(2x + 2)}{ \sqrt{1 - { ({x}^{2} + 2x)}^{2} } }$

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