G
Guest
Guest
Please can you check this for me?
Use the Product Rule to differentiate the function
\(\displaystyle \L k(x) = (\cos x)e^{ - x\sqrt 3 } \\)
I applied the following method:
\(\displaystyle \L k'(x) = f'(x)g(x) + f(x)g'(x) \\)
where
\(\displaystyle \L f(x) = \cos x{\rm }f'(x) = - \sin x \\)
\(\displaystyle \L g(x) = e^{ - x\sqrt 3 } {\rm }g'(x) = - \sqrt 3 .e^{ - x\sqrt 3 } \\)
which gave
\(\displaystyle \L k'(x) = e^{ - x\sqrt 3 } ( - \sin x) - \sqrt 3 .e^{ - x\sqrt 3 } (\cos x) \\)
\(\displaystyle \L = e^{ - x\sqrt 3 } ( - \sin x - \sqrt 3 \cos x) \\)
\(\displaystyle \L = - e^{ - x\sqrt 3 } (\sin x + \sqrt 3 \cos x) \\)
The problem is that I have been advised the answer should be
\(\displaystyle \L e^{ - x\sqrt 3 } (\sin x + \sqrt 3 \cos x) \\)
that is, without the minus sign at the start of the expression.
Please can you explain how to get rid of the minus sign?
Thank you
:?:
Use the Product Rule to differentiate the function
\(\displaystyle \L k(x) = (\cos x)e^{ - x\sqrt 3 } \\)
I applied the following method:
\(\displaystyle \L k'(x) = f'(x)g(x) + f(x)g'(x) \\)
where
\(\displaystyle \L f(x) = \cos x{\rm }f'(x) = - \sin x \\)
\(\displaystyle \L g(x) = e^{ - x\sqrt 3 } {\rm }g'(x) = - \sqrt 3 .e^{ - x\sqrt 3 } \\)
which gave
\(\displaystyle \L k'(x) = e^{ - x\sqrt 3 } ( - \sin x) - \sqrt 3 .e^{ - x\sqrt 3 } (\cos x) \\)
\(\displaystyle \L = e^{ - x\sqrt 3 } ( - \sin x - \sqrt 3 \cos x) \\)
\(\displaystyle \L = - e^{ - x\sqrt 3 } (\sin x + \sqrt 3 \cos x) \\)
The problem is that I have been advised the answer should be
\(\displaystyle \L e^{ - x\sqrt 3 } (\sin x + \sqrt 3 \cos x) \\)
that is, without the minus sign at the start of the expression.
Please can you explain how to get rid of the minus sign?
Thank you
:?:
