derivatives

[ineedhelp123]

find the derivative.


#1.)

f(x) = (3x+1)^3/(x)^6


#2.)

f(x) = 1/2x+sin^3x


..if someone could help me out Fast! id really appreciate it. thanks

 

[tkhunny]

You are going to have to demonstrate where you are and why you are struggling.

Have you met the Quotient Rule and the Chain Rule? We can't get very far without them, since I'm confident you do not want to find the derivatives usign the definition only.

Let's see what you get.

 

[Unco]

#1.)

f(x) = (3x+1)^3/(x)^6

\(\displaystyle \mbox{ }\)Firstly, by the chain rule:

\(\displaystyle \mbox{ \frac{d}{dx} (3x + 1)^3 = 3(3x + 1)^2 \cdot 3 = 9(3x + 1)^2}\)

You could use the quotient rule:

\(\displaystyle \L \mbox{ \frac{d}{dx} \frac{(3x + 1)^3}{x^6} = \frac{9(3x + 1)^2 \cdot x^6 - 6x^5 \cdot (3x + 1)^3}{x^{12}}}\)

\(\displaystyle \mbox{ }\)and simplify down.

Or you could use the product rule:

\(\displaystyle \mbox{ \frac{d}{dx} \frac{(3x + 1)^3}{x^6} = \frac{d}{dx} (3x + 1)^3x^{-6} = etc}\)

#2.)

f(x) = 1/2x+sin^3x

It's not clear what you mean here.

If it's f(x) = 1/(2x + sin^3(x)),

You could write as

\(\displaystyle \mbox{ \left(2x + \sin^3{(x)}\right)^{-1}}\) and practise your chain rule.

\(\displaystyle \mbox{ }\)Ok, by the chain rule:

\(\displaystyle \mbox{ \frac{d}{dx} \sin^3{(x)} = \frac{d}{dx} \left(sin{(x)}\right)^3 = 3\left(sin{(x)}\right)^2 \cdot \cos{(x)}}\).