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To differentiate y=sin(sin(x)) you need to use the chain rule.
A common way to remember the chain rule is "derivative of the outside, keep the inside, derivative of the inside".
First, you take the derivative of the outside. The derivative of sin is cos.
Then, you keep the inside, so you keep sin(x).
Then, you multiple by the derivative of the inside. Again, the derivative of sinx is cosx.
In the end, you get y'=cos(sin(x))cos(x))
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How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
What is sin x times sin x?
We write sin x * sin x = sin2 x
What is sin x differentiated?
The differential of sin x with respect to x is: d(sin x) = cos x dx
Why does the derivative of sin x equal - cos x?
It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.
How do you factor sin squared times x plus cos2x -cosx equals 0?
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