Find derivative with chain rule

[pavadrin]

hey
how would i find the derivative of y= -18 \sin 80 t?
thanks pavadrin

 

[HallsofIvy]

1. Use the fact, which you should already have learned, that the derivative of sin(x) is cos(x).
2. Use the fact, which you should already have learned, that the derivative of 80t is 80.
3. Use the chain rule.

By the way, the derivative is one of the basic operations in calculus so this is clearly not "pre-calculus". I'm moving this to "Calculus and beyond".

 

[pavadrin]

okay sorry for having placed it in the wrong section, um...ive never differentiated a trig f(x) before that's all and it was the first i came across that i needed to differentiate. could i please given a few hints on the basics? thanks

 

[Astronuc]

I would imagine if one is studying differential calculus, that one is using a textbook and within the textbook are trigonometric identities and examples.

HallsofIvy mentioned the chain rule, which is very basic in differentiation.

Given y = f(g(x)), y' = dy/dx = f'(g(x))*g'(x).

Has one proved to oneself the definition of a derivative, and then used that definition to find the derivative of various functions?

y'(x) = dy(x)/dx = \lim_{\substack{\Delta{x}\rightarrow 0}} \frac{y(x+\Delta{x})-y(x)}{\Delta{x}}

http://hyperphysics.phy-astr.gsu.edu/hbase/deriv.html
http://hyperphysics.phy-astr.gsu.edu/hbase/math/derfunc.html#c1

http://en.wikipedia.org/wiki/Derivative
http://en.wikipedia.org/wiki/Chain_rule

http://mathworld.wolfram.com/Derivative.html

 

[pavadrin]

okay thanks for the replies and the links. so the derivative of sin (x) = cos (x) therefore that if y = -18sin (80t) then y' = 80(-18cos (80t)), expanding the brackets is equal to y' = -1440cox 80t? thanks

 

[VietDao29]

okay thanks for the replies and the links. so the derivative of sin (x) = cos (x) therefore that if y = -18sin (80t) then y' = 80(-18cos (80t)), expanding the brackets is equal to y' = -1440cox 80t? thanks

Yup, this is correct.
A harmless typo though, cox should read cos, instead. :)

 

[pavadrin]

oh okay thanks for confirming that