# Given g(x)=cosx+xsin x, 0<x<pi/2, verify if g'(x)>0? explain You need to differentiate the given function with respect to `x` , such that:

`g'(x) = (cos x + x*sin x)'`

`g'(x) = (cos x)' + (x*sin x)'`

You need to differentiate the product `x*sin x` using the product rule, such that:

`g'(x) = (cos x)' + x'*sin x +...

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You need to differentiate the given function with respect to `x` , such that:

`g'(x) = (cos x + x*sin x)'`

`g'(x) = (cos x)' + (x*sin x)'`

You need to differentiate the product `x*sin x` using the product rule, such that:

`g'(x) = (cos x)' + x'*sin x + x*(sin x)'`

`g'(x) = -sin x + sin x + x*cos x`

Reducing duplicate terms, yields:

`g'(x) = x*cos x`

You need to prove that `g'(x) > 0` , hence, you need to prove either both factors, `x` and `cos x` , are positive, or both factors are negative.

Since `x in (0,pi/2)` , hence, `x>0` , you need to prove that `cos x>0` for `x in (0,pi/2)` .

You need to remember that the values of `cos x` are positive in quadrant 1 (`x in (0,pi/2)` ), hence, the factor `cos x > 0.`

Hence, since the function `g'(x)` is a product of two positive factors, yields that `g'(x)>0` for `x in (0,pi/2).`