How do you find the derivative of $s=(2t+1)(3t-2)$ ?

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Answer 1 · 2017-06-01T22:46:36

Answer: $s'=12t-1$

Differentiate $s=(2t+1)(3t-2)$

Consider the product rule:
$h(x)=f(x)*g(x)$
$h'(x)=f'(x)*g(x)+f(x)g'(x)$

So, we have that:
$s'=(2)(3t-2)+(2t+1)(3)$
$s'=6t-4+6t+3$
$s'=12t-1$

Answer 2 · 2017-06-02T00:25:09

$(ds)/dt = 12t - 1$

The other answer is completely valid, I just thought I would show the other method of differentiation in this case.

First, use the FOIL method (from all the way back in Algebra 1) to expand the polynomial.

$s = (2t+1)(3t-2)$

$s = 6t^2 - 4t +3t - 2$

$s = 6t^2 - t - 2$

Now, differentiate using power rule. No product rule required!

$(ds)/dt = 12t - 1$

Final Answer

How do you find the derivative of s=(2t+1)(3t-2)s=(2t+1)(3t-2)?