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