
David M. answered 12/03/20
Understanding how calculus works.
Use the following rules starting with the first one and then using the subsequent rules as applicable.
Product rule of derivatives for (t2)•(e5-x)
If h(x)=[f(x)]•[g(x)],
then h’(x)=[f(x)]•[g’(x)]+[f’(x)]•[g(x)]
Power rule of derivatives for t2
If f(x)=xn, then f’(x)=nxn-1
Product [or quotient] rule of exponents (not of derivatives) for e5-x
bn•bm=bn+m
bn/bm=bn-m
Product [or quotient] rule of derivatives for e5-x
• If h(x)=[f(x)]•[g(x)] then
h’(x)=[f(x)]•[g’(x)]+[f’(x)]•[g(x)]
• If h(x)=[f(x)] / [g(x)], then
h’(x)={[g(x)]•[f’(x)]-[f(x)]•[g’(x)]} / {[g(x)]2}
Broad A.
Thanks for your reply. I think I have already figured it out, one would have to apply both the product rule and the chain rule for the second part, so:A'(t) = 2t * e^(5-t) + t^2(-e^(5-t))
11/26/20