Please help asap. thank you

For B i am getting (8tsqrtx+8tx+2tsqr-3)/2sqrtx

For C i am getting (-tsqr-1)/2t^3/2)

For D i am getting (2u-1)/(3u(u-1)^2/3))
 
djdownfawl said:
For B i am getting (8tsqrtx+8tx+2tsqr-3)/2sqrtx

For part (B), y is a function of x, not t.

Treat t^2 as a constant, and derivate with respect to the variable x.
Click to expand...

It's not necessary, but I would multiply the factor (1 + sqrt(x)) by the number [2t^2 - 3] first.

p(x) = [2t^2 - 3] + [2t^2 - 3] sqrt(x)

Treat the bracketed expression as a constant.

I mean, if the value 2t^2 - 3 were to be 5, for example, how would you determine the derivative of this?

5 + 5x^(1/2)

It's 0 + 5/2 x^(-1/2), yes ?

Works the same way, writing 2t^2 - 3 instead of 5.

MY EDIT: fixed typographical error, and changed name from "expression" to "factor"
 
djdownfawl said:
For C i am getting (-tsqr-1)/2t^3/2)

We can't write sqrt(-1), and you're missing some grouping symbols.

Please proofread your posts, before submitting.
Click to expand...

I get u`(t) = 3/2 t^(1/2) + 1/2 t^(-3/2)

 
djdownfawl said:
I am not able to understand you

That's because my first reply contained a typographical error. I fixed it.

Please excuse me.
Click to expand...

Do you know how to determine the derivative of the following:

c + c * sqrt(x)

where c is an arbitrary constant and x is the variable ?

 
djdownfawl said:
For D i am getting (2u-1)/(3u(u-1)^2/3))

Hmmm, seems sorta close.
Click to expand...

Can you start showing your work ? (That makes my life easier.)

I get 1/3 (2u - 1) (u^2 - u)^(-2/3)
 


Hi. I have time now only to look at one exercise; I'll come back later tonight sometime.

On exercise (B), we want the derivative with respect to x, yes?

p`(x) = (2t^2 - 3)/[2 sqrt(x)]

You set up the product rule properly, but (2t^2- 3) is treated as a constant, when examining the change in y with respect to x.

The derivative of 2t^2 - 3 is therefore zero, with respect to x, because the variability of x has no effect on the value of 2t^2 - 3.

When x changes value, only the expressions containing x change value.

So, that zero eliminates the first half of the product-rule setup. You're left with the following.

p`(x) = (2t^2 - 3) x^(-1/2)/2

Some people do not like negative exponents, so we could rewrite it for them.

p`(x) = (2t^2 - 3)/[2 sqrt(x)]

 
problem c:

\(\displaystyle \frac{t^2 \ + \ 1}{\sqrt{t}}\)

\(\displaystyle = \ t^{\frac{3}{2}} \ + \ t^{- \ \frac{1}{2}}\)

now continue....
 
Please help me with these two problems i am very close to the answer where am i WRONGGG???
 
For problem E, you took the derivative wrong.

You are using the chain rule, and the first part is right, but when you take the derivative of the second where you have:

u/v

You took (v'u - uv')/v^2

You must first take u' times v - v' times u...so it should look like:

(u'v - v'u)/v^2.

Instead of getting x - 1 - x - 1 you should have gotten x + 1 - x + 1.

So you would have a positive 2 rather than a negative 2. The end result simply changes sign!

Hope that helps with problem E. :D
 
You are going wrong in the last step:

\(\displaystyle = \frac{3}{2}*t^{\frac{1}{2}} \ - \ \frac{1}{2}*t^{- \ \frac{3}{2}}\)

\(\displaystyle = \frac{3*t^2 - 1}{2*t^{\frac{3}{2}}}\)
 
I think i am right on problem E .. i double checked.
Can anybody confirm please.
Thanks
 
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