Product rule to differentiate please help (calculus)?

Question

3 on sqrt(t) (t²+4)

i cant do that script on the keyboard but it looks something like this ⁿ√(t) (t²+4) except n is a 3...

lol ty pascal! lol.... cube root.. brain no function well today

Answer 1

You mean

(t)^(1/3)*(t^2+4)

d(t)^(1/3) = (1/3)t^(1/3-1) = (1/3)t^(-2/3)

d(t^2+4) = 2t

d[(t^(1/3)(t^2+4)] = (t)^(1/3)*d(t^2+4) + (t^2+4)*d(t)^(1/3)

=>(t)^(1/3)*(2t) + (t^2+4)*(1/3)(t)^(-2/3)

=>2t^(1+1/3) + (1/3)*(t^(2-2/3) + (4/3)t(-2/3)

=>2t^(4/3) + (1/3)[t^(4/3) + 4/3 t^(2/3)]

=>[6t^(4/3+2/3) + t^(4/3 + 2/3) + 4)]/3t^(2/3)

=>[6t^2 + t^2 + 4]/(t^(2/3)

=>(7t^2+4)/t^(2/3)

Answer 2

Edit: you're welcome. Hope your brain functions better tomorrow.

Oi vei. I can understand that not everyone knows where to find the unicode character ∛, but you should know what function it refers to. It's CUBE root, not "square root with a 3." Acceptable abbreviations for ∛x are cbrt(x) and x^(1/3).

Moving on...

d(∛t (t²+4))/dt
d(∛t)/dt (t²+4) + ∛t d(t²+4)/dt
d(t^(1/3))/dt (t²+4) + ∛t d(t²+4)/dt
1/3 t^(-2/3) (t²+4) + ∛t · 2t
(t² + 4)/(3∛(t²)) + 2t∛t

And we are done.

Answer 3

(7t^2+4)/3t^(2\3) if the cube root is only on the first t

Answer 4

Is the root over the whole expression? if so take the root and make it a exponent (1/3) then do your chain rule. If its only on the first t just go

f(t)=g(t)h(t)

f'(t)=g'(t)h(t)+g(t)h'(t)