differentiation question

mathhelp1a

New member
Joined
Oct 4, 2009
Messages
40
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 )
 
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

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


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

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

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