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

2 Answers
Feb 11, 2018

Answer:

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#

Feb 11, 2018

Answer:

#(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))#