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differentiation of parametric equation

Hello guys I'm trying to differentiate a parametric equation which i attached you here but there's a part i don't understand when it comes down to differentiate y=2tcost and how it was differentiated . i know that differentiation of cos t= -sin t hence dy/dt of y=2tcost should be given as -2tsint am i missing something here?
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Reply 1
Avatar for Zacken
Zacken
22
Original post by Alen.m
Hello guys I'm trying to differentiate a parametric equation which i attached you here but there's a part i don't understand when it comes down to differentiate y=2tcost and how it was differentiated . i know that differentiation of cos t= -sin t hence dy/dt of y=2tcost should be given as -2tsint am i missing something here?


You need to use the product rule since t is a variable. ddt(2tcos⁡t)≠2tddt(sin⁡t)\frac{d}{dt} (2t \cos t) \neq 2t \frac{d}{dt}(\sin t)
Reply 2
Avatar for Zacken
Zacken
22
Original post by Alen.m
...


Surely you wouldn't agree with be saying ddx(x⋅x)=xddx(x)=x⋅1=x\frac{d}{dx} (x \cdot x) = x \frac{d}{dx} (x) = x \cdot 1 = x, riiiight?
Reply 3
Avatar for Alen.m
Alen.m
OP
11
Original post by Zacken
Surely you wouldn't agree with be saying ddx(x⋅x)=xddx(x)=x⋅1=x\frac{d}{dx} (x \cdot x) = x \frac{d}{dx} (x) = x \cdot 1 = x, riiiight?


course not :smile:)) so basically when there are two functions multiplies together we use product rule
Reply 4
Avatar for Zacken
Zacken
22
Original post by Alen.m
course not :smile:)) so basically when there are two functions multiplies together we use product rule


Yes.
Reply 5
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Alen.m
OP
11
Original post by Zacken
Yes.


thanks :smile:

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