what is the derivitive of y=sin(sin(cosx)).?

Question

can u pls explain also... thank you...

Answer 1

According to the chain rule, the derivative of the sine of something is the cosine of that something times the derivative of that something:

y' = cos(sin(cos x))*(derivative of something)
y' = cos(sin(cos x))*(sin(cos x))'
y' = cos(sin(cos x))*cos(cos x)*-sin x

Answer 2

Start on the inside and let u = cos x and then let v = sin u and so, the problem becomes:
y = sin v
y' = cos v * (v') = cos v *(cos u)*(u') = cos v * cos u * (-sin x)
= cos(sin(cosx)) *cos(cos x)*(-sin x)
= - sin x * cos(sin(cos x))*cos(cos x)

Answer 3

In general sin(something) when diffentiated gives cos(something)* (something)' so just repeat the pattern .anyway you have detailed answers by some friends:)

Answer 4

First let u = sin(cos(x)), then y = sin(u)
Use the chain rule
dy/dx = dy/du * du/dx =

cos(u) * d/dx sin(cos(x));

now let v = cos(x); use the chain rule again:

d/dx sin(v) = d/dv(sin(v) * dv/dx = cos(v) * dv/dx,

dv/dx = d(cos(x)/dx = -sin(x).

Put it all together

cos(u) * cos(v) * (-sin(x)); putting in for u and v:

cos(sin(cos(x))) * cos(cos(x)) * (-sin(x))

Answer 5

ans:
-cos(cosx(-sinx))=cos(cos(sinx))