Question

What is the derivative of y=3sin8x ?

Answer 1

apply the following rule

\[\frac{ d }{ dx } \sin(f(x)) = f'(x) \cos(f(x))\]

Answer 2

Also known as chain rule, of function of a function rule.

The full rule goes
\[\frac{ d }{ dx } f(g(x)) = f'(g(x)) g'(x)\]
or \[\frac{ dy }{ dx } = \frac{ dy }{ du } \times \frac{ du }{ dx }\]

Answer 3

\[3\cos(8x)*8\]
\[24\cos(8x)\]

Answer 4

So you multiply the coefficient of the variable x by the constant in front of sin?

Answer 5

yes but don't forget you need to also multiply by the directive of sin for that function.

So d/dx sin(f(x)) -->f'(x) cos(f(x)) where f(x) = 8x, so you get f'(x) = 8 and
8 cos (8x)

Answer 6

For this question, yes. But it can be different for different questions.

Answer 7

sorry should say 24 cos(8x) as you multiply original by 3.

Answer 8

My teacher told me the chain rule as jump-in rule. so I used that. If your are interested in learning that way, let me know.

Answer 9

In fact it might help you see this pattern for composite functions of trigonometric functions.

\[\frac{ d }{ dx } \sin x = \cos x ; \frac{ d }{ dx } \sin (f(x)) = f'(x) \cos (f(x))\]
\[\frac{ d }{ dx } \cos x = -\sin x ; \frac{ d }{ dx } \cos (f(x)) = -f'(x) sin (f(x))\]
\[\frac{ d }{ dx } \tan x = \sec^2 x ; \frac{ d }{ dx } tan (f(x)) = f'(x) \sec^2(f(x))\]

Answer 10

Oooh okie dokie now I'm remembering it all, thank you both a heap

Answer 11

Factor out the constant 3
Let u=8x, d/du= cos(u)

Using the the chain rule from above.

Answer 12

Okay so then u' or du/dx=8 so it gets brought to the front and multiplied by the 3 in the final equation?

Answer 13

chain rule

Answer 14

outside ' ( inside (inside )) * ( inside ' (inside )) * (inside )'

Answer 15

Not quite the final yet, but yes. Then you get 24 cos(8x) (d/dx (x))