Question

Differentiate the following function. g(t) = 5t3 cos(t)?

Answer 1

This is the derivative of a product of functions, therefore the result comes from f'g+g'f where f=5t3 and g=cos(t),

f' is the derivative of f which results from multiplying the coeficient (5) by the exponential (3) which equals 15 and substracting 1 from the exponential. Therefore f'=15t2

g' is the derivative of the function cos(t) which is -sin(t). This we are supposed to know by heart.

So, f'g+g'f= 15t2 cos(t) + (-sin(t) 5t3 )

By multiplying the signs you get: 15t2 cos(t) - sin(t) 5t3

If you use the common factor procedure you get 5t2, the common factor is the biggest number by which you can divide every coefficient in the expression. It also is the biggest exponential you can substract from the expressions that have exponentials. In this case is 5 for the coefficients and t2 for the exponentials. You put the common factors outside the expression multiplying it by the rest.

The result is= 5t2 (3 cos(t) - sin(t) t)