More Derivatives

• Apr 8th 2006, 07:44 PM
Hallah_az
More Derivatives

f(x)=1/x

k(x)=2x/x^2+2

v(x)=(x^2+3)^

m(x)=-5(x^4+3)^8
• Apr 8th 2006, 10:15 PM
earboth
Quote:

Originally Posted by Hallah_az

f(x)=1/x
k(x)=2x/x^2+2
v(x)=(x^2+3)^
m(x)=-5(x^4+3)^8

Hello,

to 1.:
$\displaystyle f(x)=x^{-1} \Longrightarrow \frac{df}{dx}=(-1) \cdot x^{-2}=- \frac{1}{x^2}$

to 2.: You have to use the quotient rule, if the function reads like this:
$\displaystyle k(x)=\frac{2x}{x^2+2}$

$\displaystyle \frac{dk}{dx}=\frac{(x^2+2) \cdot 2-2x \cdot 2x}{(x^2+2)^2} =\frac{-2x^2+4}{(x^2+2)^2}$

to 3.: Your problem is not complete. So I'll use n as exponent. You have to plug in the right value. You have to use the chain rule:

$\displaystyle v(x)=(x^2+3)^n \longrightarrow \frac{dv}{dx}=n \cdot (x^2+3)^{n-1} \cdot 2x$

to 4.: You have to use the chain rule:

$\displaystyle m(x)=-5(x^4+3)^8 \Longrightarrow \frac{dm}{dx}=(-5) \cdot 8 \cdot (x^4+3)^7 \cdot 4x^3$

$\displaystyle \frac{dm}{dx}=-160x^3(x^4+3)^7$

Greetings

EB