How do you find the derivative of #f(x)=2/root3(x)+3cosx#?

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Answer 1 · 2017-12-03T18:21:03

#-2/(3root(3)x^4)-3sinx#

First, we can write those square root/ other roots into a simple form #x^n#.
Then we can apply the power rule:
#y=x^n#
#dy/dx=y'=nx^(n-1)#

Let's start to solve it!
#f(x)=2/root(3)x+3cosx#
#=2x^(-1/3)+3cosx#

#f'(x)=2(-1/3)x^(-1/3-1)+3(-sinx)#
#=-2/3x^(-4/3)-3sinx#
#=-2/(3root(3)x^4)-3sinx#

Here is the answer. Hope it can help you :)
Feel free to ask me if you have any questions.

Answer 2 · 2017-12-03T18:22:38

Please see below.

Use the power rule and the derivative of cosine.

#d/dx(2/root(3)x) = d/dx(2x^(-1/3)) = -2/3x^(-4/3) = -2/(3x^(4/3))#

#d/dx(3cosx) = 3(-sinx) = -3sinx#.

So,

#f'(x) = -2/3x^(-4/3)-3sinx#

Rewrite the first term to preference.