Whats the derivative of: sin(squared)x + cosx ?

Question

sin^2x + cosx or (sinx)^2 + cosx

Answer 1

Use the chain rule.

d/dx [(sin(x))^2 + cos(x)] =

2 * sin(x) * cos(x) - sin(x)
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Note added in response to answerers who don't think the chain rule applies here.

The chain rule most definitely applies. For the first term of the expression in question, we have:

(sin(x))^2

Let f(x) = sin(x) and let g(y) = y^2. The first term in the expression can therefore be written as:

g(f(x))

This is a compound function, the derivative of which is given by:

g'(f(x))*f'(x)

where the "prime" marks represent derivatives.

In this case, f'(x) = cos(x) and g'(f(x)) = 2f(x) = 2*sin(x), so the derivative of the first term is given by:

2*sin(x)*cos(x)

Answer 2

I take it that the sin is squared (not the x).
Then you need to use that (f^2)' = 2*f*f' and you get: 2*sin(x)*cos(x)-sin(x). You can also rewrite 2sin(x)cos(x) as sin(2x).

Answer 3

you dont need the chain rule for this problem, the first answer is right

Answer 4

2Cosx-Sin x