Solving Quick Derivative: Chain & Product Rules

[ziddy83]

hey guys,

I need to find the derivative of this function, do i use the chain rule and the product rule? and for the stuff inside the parenthesis, how do i differentiate that? the derivative of x is just 1...but since its a fraction, would it be as simple as that? please help, thanks.

\frac{1}{32} \left( \frac{64}{x} + \frac{x}{50} \right) * 1.60

 

[dextercioby]

hey guys,

I need to find the derivative of this function, do i use the chain rule and the product rule? and for the stuff inside the parenthesis, how do i differentiate that? the derivative of x is just 1...but since its a fraction, would it be as simple as that? please help, thanks.

\frac{1}{32} \left( \frac{64}{x} + \frac{x}{50} \right) * 1.60

Write it as a sum:3.2x+\frac{1}{1000x} and use the derivatives of "x" and "1/x" to find your result.

 

[quasar987]

mmh.. all you need is the property [ k*f(x) ]' = k*f '(x) for any constant k.

In your case,

\frac{1}{32} \left( \frac{64}{x} + \frac{x}{50} \right) * 1.60 = \frac{1.60}{32}\left( \frac{64}{x} + \frac{x}{50} \right) = \frac{1.60}{32}\left( \frac{64}{x} \right) + \frac{1.60}{32}\left(\frac{x}{50} \right) = \frac{1.60*64}{32}\left( \frac{1}{x} \right) + \frac{1.60}{32*50}\left(x \right)

and 1/x is the same as x^{-1}. And you know what the derivative rule is for x^k[/itex] where k is a constant.

 

[ziddy83]

Great, thanks a lot guys, i appreciate it.