What is the derivative of #g(x)=x+(4/x)#?

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Answer 1 · 2018-07-30T20:24:21

#g'(x) = 1-4/(x^2)#

To find the derivative of #g(x)#, you must differentiate each term in the sum

#g'(x) = d/dx(x) + d/dx(4/x)#

It is easier to see the Power Rule on the second term by rewriting it as

#g'(x) = d/dx(x) + d/dx(4x^-1)#

#g'(x) = 1 + 4d/dx(x^-1)#

#g'(x) = 1 + 4(-1x^(-1-1))#

#g'(x) = 1 + 4(-x^(-2))#

#g'(x) = 1 - 4x^-2#

Finally, you can rewrite this new second term as a fraction:

#g'(x) = 1-4/(x^2)#

Answer 2 · 2018-07-30T20:30:35

#g'(x)=1-4/(x^2)#

What might be daunting is the #4/x#. Luckily, we can rewrite this as #4x^-1#. Now, we have the following:

#d/dx (x+4x^-1)#

We can use the Power Rule here. The exponent comes out front, and the power gets decremented by one. We now have

#g'(x)=1-4x^-2#, which can be rewritten as

#g'(x)=1-4/(x^2)#

Hope this helps!