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# Help with derivatives question?

What is f'(x) if f(x) = cos x^2

### 8 Answers

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• Since no one actually explained it, I'll do that, here.

Presuming that is:

f(x) = cos(x²)

We can set up the chain rule:

f = cos(u) and u = x²

Get the derivative of both:

df/du = -sin(u) and du/dx = 2x

Now we can apply the chain rule:

df/dx = df/du * du/dx

df/dx = -sin(u) * 2x

df/dx = -2x sin(u)

df/dx = -2x sin(x²)

Back into function notation:

f'(x) = -2x sin(x²)

• f(x)=cos(x^2)

=>

f '(x)=-sin(x^2)(2x)

=>

f '(x)=-2xsin(x^2)

• f(x) = cos(x^2)

f'(x) = - 2xsin(x^2) answer//

• d/dx(cos(x^2)) = -2 x sin(x^2)

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• f'(x) = -2x sin x^2

• First way:

f(x) = cos²(x)

Recall the derivative of u^(n) is: n * u' * n^(n - 1) → where in your case:

u = cos(x) → u' = - sin(x)

n = 2

f'(x) = n * u' * n^(n - 1)

f'(x) = 2 * [- sin(x)] * [cos(x)]^(2 - 1)

f'(x) = - 2.sin(x).cos(x)

Second way:

f(x) = cos(x) * cos(u)

Recall the derivative of u^(n) is: (u'.v) + (v'.u) → where in your case:

u = cos(x) → u' = - sin(x)

v = cos(x) → v' = - sin(x)

f'(x) = (u'.v) + (v'.u)

f'(x) = [- sin(x) * cos(x)] + [- sin(x) * cos(x)]

f'(x) = - sin(x).cos(x) - sin(x).cos(x)

f'(x) = - 2.sin(x).cos(x)

• I think you mean f(x) = cos²x

Using the chain rule we can show that:

f '(x) = -2cosxsinx

Now, 2cosxsinx => sin2x

so, f '(x) = -sin2x

:)>

• f'(x)= -2cosx*sinx

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