I can't do derivative of:(1025t^2-800t+400)^(1/2)

[thomasrules]

I can't do derivative of:

(1025t^2-800t+400)^(1/2)

 

[dextercioby]

Do you know the chain rule??If so,apply it.

Daniel.

 

[vincentchan]

use chain rule...
the derivative of x^1/2 = 1/2 x^-1/2

chain rule and above formulas should solve your problem... if you have no idea how to apply chain rule, look it up in your textbook and study carefully, you must fully understand before you goto the next chapter, otherwise, you will have a tough time in the future.

 

[thomasrules]

thanks you :)

ya I am doing Optimization in Economics and Science...SO DAMN HARD

so I end up getting f1(x)=(1/2)(1025t^2-800t+400)^(-1/2)(2050t-800)

THen my teacher jumps too: If f1(x)=0 2050t=800

WTF?

 

[Zurtex]

I asusme you want to differentiate with respect to t.

Let:

$$u=1025t^2-800t+400$$

Then your function is:

$$u^{\frac{1}{2}}$$

By the chain rule:

$$\frac{d \left( u^{\frac{1}{2}} \right)}{dt} = \frac{1}{2} \frac{du}{dt} u^{- \frac{1}{2}}$$

You should be able to take it from there.

 

[thomasrules]

thanks you :)

ya I am doing Optimization in Economics and Science...SO DAMN HARD

so I end up getting f1(x)=(1/2)(1025t^2-800t+400)^(-1/2)(2050t-800)

THen my teacher jumps too: If f1(x)=0 2050t=800

WTF?

 

[Nylex]

thanks you :)

ya I am doing Optimization in Economics and Science...SO DAMN HARD

so I end up getting f1(x)=(1/2)(1025t^2-800t+400)^(-1/2)(2050t-800)

THen my teacher jumps too: If f1(x)=0 2050t=800

WTF?

He/she is finding turning points?

 

[thomasrules]

trying to find t, sets 0 to find max or min

 

[dextercioby]

Not really,u should compute the second derivative as well.It's the only way you can decide which is max and which is min,if any...

Daniel.