# Let `f(x)= 3cos(x) *sin^(-1)(x)` Find `f'(x)` .

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### 1 Answer

`f(x) = 3cos(x)*sin^(-1)(x)`

To determine f'(x), use the product formula of derivatives which is `(uv)' = uv'+u'v` .

So let,

`u=3cos(x) ` and `v=sin^(-1) (x)`

To determine u', use the formula `d/(du)(cosu) =-sin u * u'` .

And for v', use `d/(du) sin^(-1)u = 1/(sqrt(1-u^))*u'` .

We then have,

`u'=-3sin(x) * 1` and `v'=1/sqrt(1-x^2)*1`

`u'=-3sin(x)` `v'=1/sqrt(1-x^2)`

Substitute u,v, u' and v' to the product formula of derivatives.

`f'(x)=3cos(x)*1/sqrt(1-x^2) + (-3sin(x))*sin^(-1)(x)`

`f'(x)=(3cos(x))/sqrt(1-x^2) - 3sin(x) *sin^(-1)(x)`

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**Hence, the derivative of `f(x)=3cos(x)*sin^(-1)(x)` is `f'(x)= (3cos(x))/sqrt(1-x^2) - 3sin(x)*sin^(-1)(x)` .**

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