Differentiating using a product rule

[sozener1]

when differentiating

e^(at) * (cos(bt) + isin(bt))

are you able to use product rule to find the derivative considering (cos(bt) + sin(bt)) as one function??

why??

and what does d/dt exactly mean?? (they get multiplied to a function that needs to be differentiated and I wanted to solidify their definitions in my head so that I could manipulate them when encountered)

 

[SteamKing]

when differentiating

e^(at) * (cos(bt) + isin(bt))

are you able to use product rule to find the derivative considering (cos(bt) + sin(bt)) as one function??

Yes. But according to the product rule, you also have to take the derivative of (cos(bt) + isin(bt))
w.r.t. t at some point.

why??

Because you have an exponential function multiplied by another function, which itself is the sum of two other functions. It's what the product rule is for.

and what does d/dt exactly mean?? (they get multiplied to a function that needs to be differentiated and I wanted to solidify their definitions in my head so that I could manipulate them when encountered)

You are not 'multiplying' d/dt in the arithmetic sense when you express d(some function)/dt.

'd/dt' can be thought of as an operator, like f(x) means a function of x. d/dt notation means 'take the first derivative w.r.t. the variable t.'

 

[HallsofIvy]

when differentiating

e^(at) * (cos(bt) + isin(bt))

are you able to use product rule to find the derivative considering (cos(bt) + sin(bt)) as one function??

No, you do not use the product rule to find the derivative of cos(bt)+ sin(bt) because there is NO product of two functions of t. You do need to use the "sum rule" and the "chain rule". You would need to use the product rule to differentiate the product e^(at)(cos(bt)+ isin(bt)).

why??

and what does d/dt exactly mean?? (they get multiplied to a function that needs to be differentiated and I wanted to solidify their definitions in my head so that I could manipulate them when encountered)

Any Calculus text will have a detailed explanation of "d/dt". Have you read the definition in a textbook first?

No, d/dt does NOT get "multiplied to a function". d/dt is the 'differentiation operator' and is "applied to a function". The definition, again found in any Calculus textbook, is
\frac{d}{dt} f(t)= \lim_{h\to 0}\frac{f(t+h)- f(t)}{h}.

A geometric way of looking at it is that [f(t+h)- f(h)]/h is the slope of the "secant line" to the graph of y= f(x) through (t, f(t)) and (t+ h, f(t+h)) and taking the limit as h goes to 0 gives the slope of the tangent line to the graph at (t, f(t)).