Differentiate

[nikchic5]

Differentiate the function

u= the fourth root of t^(3) + 3 * the third root of t^(4)

Thank you if anyone can help me!!!

 

[tkhunny]

It's a little tough to read, but why can't you just use the polynomial rule?

(d/dx)(x^n) = n*x^(n-1)

It works for Fractional n

 

[nikchic5]

Hey?

How else can I type in the problem? Without writing it like that? Thanks

 

[tkhunny]

The way you have it, it's not clear what you MEAN. It can be translated strictly, but that may not be your intent.

Original:
the fourth root of t^(3) + 3 * the third root of t^(4)

More Parentheses:
the fourth root of (t^(3) + 3) * the third root of (t^(4))

MORE Parentheses:
[the fourth root of (t^(3) + 3)] * [the third root of (t^(4))]


Exponential Form:
[t^3 + 3]^(1/4) + (t^4)^(1/3)

LaTeX:

\(\displaystyle \sqrt[4]{t^{3}+3}+\sqrt[3]{t^4}\)

 

[nikchic5]

I am sorry the question is


[the fourth root of (t^(3)] + [3 * the third root of (t^(4))]

 

[Gene]

I think that can be written as
[t^(3)]^(1/4) + 3 *[t^(4))]^(1/3)
=
t^(3/4)+3[t^(4/3)]
Then you can use TKH's suggestion
(d/dx)(x^n) = n*x^(n-1)
where first n = 3/4 then n=4/3

 

[tkhunny]

I am sorry the question is
[the fourth root of (t^(3)] + [3 * the third root of (t^(4))]

See, I thought I knew what you meant so I failed to translate it strictly, which would have been correct. Better to write so that no one can misunderstand.

Glad to have you with us. What else are you working on?

 

[nikchic5]

Hey...

Im just working on Calculus right now. What about you?

Do you know how to compute that problem?

 

[tkhunny]

If I answer your last question, it will be the third time. Just use the rule we started with, way up in my first response.