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

Answer 1

#-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.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Samantha Adkison

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.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Adam Adkins

To find the derivative of ( f(x) = \frac{2}{\sqrt{3}x} + 3\cos(x) ), you can use the sum and power rules of differentiation. The derivative of ( \frac{2}{\sqrt{3}x} ) is ( -\frac{2}{\sqrt{3}x^2} ), and the derivative of ( 3\cos(x) ) is ( -3\sin(x) ). Therefore, the derivative of ( f(x) ) with respect to ( x ) is:

[ f'(x) = -\frac{2}{\sqrt{3}x^2} - 3\sin(x) ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.