Question

if y=sin(3sin-1x), then dy/dx is equal to​

Answer 1

The derivative of the given function is (3 cos ( 3si$n^{-1}$x )) / ${\sqrt{1-x^{2} }$.

Given:

y = sin( 3si$n^{-1}$x )

To Find:

The first derivative of y.

Solution:

To solve this problem, we will use the concepts given below:

i) We will apply the chain rule here. It states that if u and v are two functions such that u $o$ v = h such that u(v(x)) = h(x), then

h'(x) = u'(v(x)) . v'(x)

ii)  $\frac{d}{dx}$ (sin x) = cos x

iii)  $\frac{d}{dx}$ ( si$n^{-1}$x ) = $\frac{1}{\sqrt{1-x^{2} } }$

Our given function is y = sin( 3si$n^{-1}$x ).

Applying the chain rule,

$\frac{dy}{dx}$ =  $\frac{d}{dx}$( sin( 3si$n^{-1}$x )).  $\frac{d}{dx}$( 3si$n^{-1}$x ) =  $\frac{d}{dx}$( sin( 3si$n^{-1}$x ))× 3 $\frac{d}{dx}$( si$n^{-1}$x )

⇒ $\frac{dy}{dx}$ =  cos ( 3si$n^{-1}$x ) × $\frac{3}{\sqrt{1-x^{2} } }$ = (3 cos ( 3si$n^{-1}$x )) / ${\sqrt{1-x^{2} }$

∴ The derivative of the given function is (3 cos ( 3si$n^{-1}$x )) / ${\sqrt{1-x^{2} }$.

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