Derivative of function

[dubb]

Find the derivative of function:
2(x^n)sinx

I can't seem to find examples on this one. This is what I have tried so far.

I used the product rule.

(2nx^n-1) (cosx) + (2x^n)(sinx)
2nx^n-1 cosx + 2x^n sinx

2(nx^n-1 cosx + x^n sinx), thats as far as I can get

Thanks for your help!

 

[HallsofIvy]

Find the derivative of function:
2(x^n)sinx

I can't seem to find examples on this one. This is what I have tried so far.

I used the product rule.

(2nx^n-1) (cosx) + (2x^n)(sinx)

This is NOT the product rule. If we write this as fg with f= 2x^n and g= sin(x)
then (fg)'= f'g+ fg'= (2nx^{n-1})sin(x)+ (2x^n)cos(x)

What you have would be f'g'+ fg

2nx^n-1 cosx + 2x^n sinx

2(nx^n-1 cosx + x^n sinx), thats as far as I can get

Thanks for your help!

 

[dubb]

Ohh, I see the mistake.
2x^n sinx

f' g + f g'
2n{x^n-1} (sinx) + (2x^n) (cosx)

Thanks!

 

[lookagain]

Ohh, I see the mistake.
2x^n sinx

f' g + f g'
2n{x^n-1} (sinx) + (2x^n) (cosx) \(\displaystyle \ \ \ \ \) <-------- No, you must have grouping symbols around the first exponent.

x^n-1 means \(\displaystyle \ x^n - 1, \ \ not \ \ x^{n - 1}.\)


It wasn't for no reason that HallsofIvy put braces around that exponent in his post.