differentiation question

[mathhelp1a]

f(t) = (t^2 - 1 ) ^5/2 (t^3 + 5)

is this correct differentiation : (t^2-1 )^5/2 (3t^2) + (t^3 + 5) [ 5/2 (t^2-1) ^ 3/2 (2t)]

simplify: (t^2-1) ^ 3/2 ( t^3 +5 ) (17t^4 -3t^2 + 35t )

 

[soroban]

Hello, mathhelp1a!

$\displaystyle f(t) \:=\: (t^2 - 1)^{\frac{5}{2}}(t^3 + 5)$

$\displaystyle \text{Is this correct differentiation? }\;f'(t) \;=\;(t^2-1 )^{\frac{5}{2}}(3t^2) + (t^3 + 5)\cdot\frac{5}{2}(t^2-1)^{\frac{3}{2}}(2t)$ . . . . Yes!

$\displaystyle \text{Simplify: }\;(t^2-1)^{\frac{3}{2}}(t^3 +5 )(17t^4 -3t^2 + 35t )$ . . . . no

$\displaystyle \text{We have: }\;3t^2(t^2-1)^{\frac{5}{2}} + 5t(t^2-1)^{\frac{3}{2}}(t^3+5)$


$\displaystyle \text{Factor: }\;f'(t) \;=\;t(t^2-1)^{\frac{3}{2}}\,\bigg[3t(t^2-1) + 5(t^3+5)\bigg]$

. . . . . . . . . .$\displaystyle = \;t(t^2-1)^{\frac{3}{2}}\,\bigg[3t^3 - 3t + 5t^3 + 25\bigg]$

. . . . . . . . . .$\displaystyle = \;t(t^2-1)^{\frac{3}{2}}(8t^3 - 3t + 25)$

 

[mathhelp1a]

can you explain the rule that you used
how did you get t in t (t^2...

also where did you get the 3t and 5