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What is the derivative of #y = sin(sin(sinx))#?

Answer 1

Samantha Adkison

We have #(sin(u))' = u'cos(u)#

where #u = sin(sin(x))#

#u' = (sin(sin(x)))' = cos(x)*cos(sin(x))#

So #(sin(u))' = cos(x)cos(sin(x))cos(sin(sin(x))#

Sorry if in your countries you need to use the "chain rules" in France we don't use that, we just remember that :

#(f(g(x)))' = g'(x)*f'(g(x))#

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Answer 2

Elijah Abram

To find the derivative of ( y = \sin(\sin(\sin(x))) ), you can use the chain rule. The chain rule states that if ( y = f(g(x)) ), then ( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) ).

Let ( u = \sin(x) ) and ( v = \sin(u) ). So, ( y = \sin(v) ).

Now, find the derivatives:

[ \frac{du}{dx} = \cos(x) ] [ \frac{dv}{du} = \cos(u) ]

Apply the chain rule:

[ \frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx} ]

[ = \cos(v) \cdot \cos(u) \cdot \cos(x) ]

Now, substitute back ( v = \sin(u) ) and ( u = \sin(x) ):

[ = \cos(\sin(\sin(x))) \cdot \cos(\sin(x)) \cdot \cos(x) ]

So, the derivative of ( y = \sin(\sin(\sin(x))) ) with respect to ( x ) is ( \cos(\sin(\sin(x))) \cdot \cos(\sin(x)) \cdot \cos(x) ).

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Answer from HIX Tutor

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