1. ## Differentiate help!

a) f(x) = (√x + (1/ x^1/3))^2
b) f(x) = √x ( 1+ sin^2 (2π/x))

2. Hello,

a) f(x) = (√x + (1/ x^1/3))^2
$\left(\sqrt{x}+\left(\frac{1}{x^{1/3}}\right)\right)^2=\left(\sqrt{x}+x^{-1/3}\right)^2$

You have something like $(u(x))^2$. Its derivative is $2u'(x)u(x)$

Here, u'(x) is the derivative of a sum.

$u'(x)=\frac{1}{2\sqrt{x}}+(-1/3)x^{-4/3}$

Hence $f'(x)=\dots$

3. Originally Posted by Rui
b) f(x) = √x ( 1+ sin^2 (2π/x))
Use the product and the chain rule

4. hello there!,

well i hope you know your basic rules of differentiation:
$\frac{d}{dx}x^n = nx^{(n-1)}$
$\frac{d}{dx}x^2 = 2x^{(2-1)} = 2x$
$\frac{d}{dx}(x+x^2)^n = n(x+x^2)^{(n-1)}\frac{d}{dx}(x+x^2)$

i will solve question a) for you and you will try to answer question b)

$
\frac{d}{dx}f(x) =\frac{d}{dx} ( x^\frac{1}{2}+ x^{- \frac{1}{3}})^2
$

$
\frac{d}{dx}f(x) =2( x^\frac{1}{2}+ x^{- \frac{1}{3}})^{(2-1)}\frac{d}{dx} ( x^\frac{1}{2}+ x^{- \frac{1}{3}})
$

$
\frac{d}{dx}f(x) =2( x^\frac{1}{2}+ x^{- \frac{1}{3}}) \left ( \frac{1}{2}x^{(\frac{1}{2} - 1)} - \frac{1}{3}x^{(- \frac{1}{3} -1)} \right)
$

$
\frac{d}{dx}f(x) =2( x^\frac{1}{2}+ x^{- \frac{1}{3}}) \left ( \frac{1}{2}x^{-\frac{1}{2} } - \frac{1}{3}x^{- \frac{4}{3} } \right)
$