Hard Trignometric Derivative Problem

[fernanhen]

Homework Statement

Find the derivative of sin(tan(square root of sinx))

Homework Equations

derivative of:

sin=cos
tan=sec squared
sinx=cosx

The Attempt at a Solution

cos(tan(square root of sinx))(sin(sec^2(1/2sinx)^-1/2(cosx))

So I did the derivative of the sin, left what's inside the parenthesis alone.

Then left sin alone, and did the chain rule inside. I'm not sure if I should also have used the product rule as well.

Help

 

[SammyS]

Homework Statement

Find the derivative of sin(tan(square root of sinx))

Homework Equations

derivative of:

sin=cos
tan=sec squared
sinx=cosx

The Attempt at a Solution

cos(tan(square root of sinx))(sin(sec^2(1/2sinx)^-1/2(cosx))

So I did the derivative of the sin, left what's inside the parenthesis alone.

[STRIKE]Then left sin alone[/STRIKE], and did the chain rule inside. I'm not sure if I should also have used the product rule as well.

Help

Drop the sine that I crossed out. There are other errors too.

$\displaystyle \frac{d}{dx}f\left(g\left(h(x)\right)\right)= f'\left(g\left(h(x)\right)\right)\cdot g'\left(h(x)\right)\cdot h'(x)$

Added in Edit:
There are four functions nested.

Therefore, $\displaystyle \frac{d}{dx}w\left(f\left(g\left(h(x)\right)\right)\right)= w'\left(f\left(g\left(h(x)\right)\right)\right) \cdot f'\left(g\left(h(x)\right)\right)\cdot g'\left(h(x)\right)\cdot h'(x)$

 

[cepheid]

Nope, you don't need the product rule, because there is no multiplication of anything by anything else here. There are no products of functions. You have a composition of functions (i.e. a function of a function etc), which means that the chain rule is exactly what you need. To help keep track of everything, it might be useful to do some substitutions. Start with the innermost function and work your way out.

Let y = sin x

let u = √y

let v = tan u

let w = sin v

So we have a composition of functions since w = w(v) = w( v(u) ) = w( v( u(y) ) )

= w( v( u( y(x) ) ) )

It's function of a function of a function of a function. So the chain rule says that:$$\frac{dw}{dx} = \frac{dw}{dv}\frac{dv}{du}\frac{du}{dy}\frac{dy}{dx}$$So all you have to do is evaluate these four derivatives separately and then multiply them together.

 

[dimension10]

Homework Statement

Find the derivative of sin(tan(square root of sinx))

Homework Equations

derivative of:

sin=cos
tan=sec squared
sinx=cosx

The Attempt at a Solution

cos(tan(square root of sinx))(sin(sec^2(1/2sinx)^-1/2(cosx))

So I did the derivative of the sin, left what's inside the parenthesis alone.

Then left sin alone, and did the chain rule inside. I'm not sure if I should also have used the product rule as well.

Help

Just do the multi-function chain rule (is that what its called?). Not the multi-variable chain rule, the multifunction chain rule.

dy/dt=dy/df df/dg dg/dh ... dp/dt