I'm having trouble approaching this problem. The problem says: differentiate with respect to t: y=bcost+t^2sint.

Question

The chain rule has been giving me problems. Maybe some tips on carrying the chain rule through? 

Answer 1

This problem uses the product rule, not the chain rule.  

 

The product rule is that if f = u*v then f' = u*v' + u'*v.  In this case u = t^2 and v = sin(t) so 

 

y' = -b*sin(t) + t^2*cos(t) + 2*t*sin(t) = (2*t - b)*sin(t) + t^2*cos(t) 

 

 

The chain rule is used when you have a composition of functions.  For example, if 

f(x) = (2x+3)^2 

then you can consider f to be a composition of the functions 

g(x) = x^2 and h(x) = 2x+3 

 

The chain rule says that if f(x) = g(h(x)) then f'(x) = g'(h(x)) * h'(x) 

so in the example above 

f'(x) = 2*(2x+3)*2 = 4*(2x+3) = 8x + 12

 

You can check this by multiplying out (2x+3)^2 and then taking the derivative term by term.  

 

Another example - say f(x) = sin(x^2)

Then f'(x) = cos(x^2) * 2x = 2x*cos(x^2)

 

Most applications of the chain rule are simple enough that you can do them in your head without having to explicitly write down what the underlying functions in the composition are.  Just keep practicing and you will get better at it!