maths q help!! (1 Viewer)

blacktown_ · Nov 21, 2012

find the second derivative of V=(t+3)(2t-1)^2

please and thankyou

CrackerMo · Nov 21, 2012

Product rule

blacktown_ · Nov 21, 2012

yeah i did it heaps of times but didn't get the right answer

Shazer2 · Nov 21, 2012

$V=(t+3)(2t-1)^{ 2 }\\ \\ Let\quad u=t+3\\ \qquad \therefore u'=1\\ Let\quad v=(2t-1)^{ 2 }\\ \qquad \therefore v'=4(2t-1)\\ \\ V'=vu'+uv'\\ \qquad =(2t-1)^{ 2 }+4(2t-1)(t+3)\\ \qquad =(2t-1)[(2t-1)+4(t+3)]\\ \qquad =(2t-1)(2t-1+4t+12)\\ \qquad =(2t-1)(6t+11)\\ \\ Let\quad u=2t-1\\ \qquad \therefore u'=2\\ Let\quad v=6t+11\\ \qquad \therefore v'=6\\ \\ V''=vu'+uv'\\ \qquad =2(6t+11)+6(2t-1)\\ \qquad =12t+22+12t-6\\ \qquad =24t+16\\ \\ \therefore \quad The\quad second\quad derivative\quad is\quad 24t+16$

That is my attempt, it might be wrong.

blacktown_ · Nov 21, 2012

omg, thankyou so much Shazer2! really appreciate it

Shadowdude · Nov 21, 2012

blacktown_ said:
yeah i did it heaps of times but didn't get the right answer

Why don't you show us your working so we can see where you went wrong

Shazer2 · Nov 21, 2012

Is it the right answer?

CrackerMo · Nov 21, 2012

blacktown_ · Nov 21, 2012

yes it's the right answer Shazer2

and Shadowdude, i will consider that for next time

maths q help!! (1 Viewer)

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