How do you find the derivative of #y=1/x-3sinx#?

1 Answer
Morgan
Mar 5, 2017

#y'=-1/x^2-3cos(x)#

Explanation:

#y=1/x-3sin(x)#

You may find it easier to rewrite this as #y=x^-1-3sin(x)#.

We can take the derivative of each term individually.

First, the derivative of #x^-1#. We can use the power rule: multiply the coefficient (in this case #1#) by the power #(-1)# , then reduce the power by one.

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

This is equivalent to #-1/x^2#.

Now we take the derivative of #3sin(x)#. Remember that when we take a derivative, we leave constants alone. The derivative of #sin(x)# is #cos(x)#. The inside term #x# would have a derivative of #1#, so the chain rule isn't necessary.

#=>d/dx(3sin(x))=3cos(x)#

So, we get:

#y'=-1/x^2-3cos(x)#

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