How do you differentiate #f(x)=sqrtxsinx#?

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Answer 1 · 2018-06-14T09:11:53

#d/dx(sqrt(x)sin(x))=sin(x)/(2sqrt(x))+sqrt(x)cos(x)#

To take the derivative of this, we need two rules. The first rule is the chain rule, which states:

#d/dx(f(x)g(x))=f'(x)g(x)+f(x)g'(x)#

The second rule we need is the power rule, which states:

#d/dx(x^n)=nx^(n-1)#

Now, we can start taking the derivative. To simplify this, I'm going to rewrite #sqrt(x)# as #x^(1/2)#, which is perfectly valid.

To start, I'm going to take the derivative of each section.

#d/dx(x^(1/2))=1/2x^(-1/2)#

#d/dx(x^(1/2))=1/(2sqrtx)#

#d/dx(sin(x))=cos(x)#

Now that we have them, we can put them together:

#d/dx(sqrt(x)sin(x))=sin(x)/(2sqrt(x))+sqrt(x)cos(x)#

And there you have your answer.