Question

How do you differentiate `(sqrt x)(x^2+3sinx)`?

Answer 1

Using differentiate product rule
`1/2(x^2+3sinx)dx/sqrtx+sqrtx(2x+3cosx)dx`
See explanation below

Explanation:

We know that in functions product, de derivative is (derivative of a product)

`(f·g)´dx=f´·g·dx+f·g´·dx`

Thus we have (if `f(x)= sqrtx` and `g(x)=x^2+3sinx`)

`f´(x)=1/2 1/(sqrtx)`

`g´(x)=2x+3cosx`.

If we apply prior formula we have

`(f·g)´dx=f´·g·dx+f·g´·dx` = `1/2(x^2+3sinx)dx/sqrtx+sqrtx(2x+3cosx)dx`

Answer 2

`(x^2+3sin(x))/(2sqrt(x))+sqrt(x)(2x+3cos(x))`

Explanation:

Product rule is `d/dx[f(x)*g(x)] =f'(x)*g(x) + f(x)*g'(x)`

Obviously we already have f(x) and g(x) so we just need to find the derivatives of both

f'(x)=`d/dx[sqrt(x)]=d/dx[x^(1/2)]=1/2x^(-1/2)=1/(2sqrt(x))`

g'(x)=`d/dx[x^2+3sin(x)]= 2x+3cos(x)`
To derive the 3sin(x) you would have to use product rule again where f(x)=3 and g(x)=sin(x), but remember the derivative of a plain number is zero, so all that's left of product rule is #f(x)*g'(x), which is just 3cos(x),remember the derivative of sin is cos.

Now just plug all the parts yo have into product rule to get:`(x^2+3sin(x))/(2sqrt(x))+sqrt(x)(2x+3cos(x))`