Question

Differentiate f(x)= square root of x times sin x

Answer 1

\[f(x)=\sqrt{x} \sin x\]

Answer 2

Use the product rule so that \[f'(x)=f(x)g'(x) + f'(x)g(x)\] when \[f(x) = \sqrt{x}\] and\[g(x) = \sin(x)\]

Answer 3

ok thanks. Is the answer cosx + 1/2 sin x?

Answer 4

No, try to solve the derivative of sqrt(x) and sinx first and then plug into the equation for the product rule.

Answer 5

the derivative of sqrt(x) is 1/2x^-1/2 correct? and the derivative of sinx is cosx..?

Answer 6

Correct

Answer 7

Now just plug that into \[f'(x)=f(x)g'(x)+f'(x)g(x)\] and simplify

Answer 8

ok I can do that but I'm not simplifying right

Answer 9

Is this your equation before you simplify?\[f'(x)=\sqrt xcos(x)+\frac{ 1 }{ 2 }x ^{-1/2}\sin(x)\]

Answer 10

\[x ^{1/2} cosx + 1/2x ^{-1/2} \sin x\]

Answer 11

Yeah :p

Answer 12

You can simplify the right side so that \[\sqrt xcos(x)=\frac{ \sin(x) }{ 2\sqrt x }\]

Answer 13

That equal sign is supposed to be a plus sign, sorry.

Answer 14

ok

Answer 15

And then you can just add the fractions or leave it like that, depending on whether the question asks you to simplify.

Answer 16

It just says differentiate

Answer 17

You didn't even have to simplify the right side then :p

Answer 18

but if it did say simplify what would be the next step.

Answer 19

Basically find the GCD which is 2sqrt(x) and multiply both sides of the left by it and add.

Answer 20

oh ok! well thank you soo much for your help!

Answer 21

No problem :)