Question

Derivative of 4 sin^2 t?

Answer 1

what have u tried?

Answer 2

I come up with 8 sin^2 t cos^2 t. using the Chain rule

Answer 3

First things first, with respect to which variable are you derivating?

Answer 4

ok,
derivative of \(y^2 \) will be 2y dy/dx
using chain rule, right ??

so derivative of \(\sin^2 t\)
will be
\(2\sin t \dfrac{d}{dt} \sin t \)
makes sense ?

Answer 5

Yes

Answer 6

so, whats the derivative of sin t ?

Answer 7

cos t

Answer 8

so whats the final answer :)

Answer 9

2 sin t cost

Answer 10

Do i multiply the 4 from the original equation by 2

Answer 11

and with the 4 in the beginning,
it'll be
8 sin t cost
or
3 sin 2t :)

oh yes :)

Answer 12

i meant 4 sin 2t ****

Answer 13

Thanks

Answer 14

i mean it s 2 sinx cosx

Answer 15

welcome ^_^

Answer 16

How did you arrive at 4 sin 2t?

Answer 17

@nelsonjedi trigonometric identity
\[\sin(2x)=2sinxcosx\]

Answer 18

the double angle formula
\(\Large 2 \sin x \cos x = \sin 2x \)

Answer 19

If you have not learned the formula yet, you can leave the answer as
\[8\times sint \times cost\]

Answer 20

However if you are doing differential calculus I assume you to be proficient with your identities....

Answer 21

Ok i am good with my original solution which I have to square But in checking my answer they show that
(8 sin t cost) ^ 2 as being 16 sin t^2 cos t^2...I found it to be 64 sin t^2 cos ^ 2...

Answer 22

Proficient and remembering are two different things...

Answer 23

you have to square your answer?

then it'll be
\(64 \sin^2 t \cos^2 t\)

Answer 24

That is what I thought..Thank you again.

Answer 25

welcome ^_^