What is the (sin(x^3-4x+2))^' ?

The (sin(x^3-4x+2))^' is cos(x^3-4x+2)(3x^2-4)

What is the first (sin(x^3-4x+2))^' ?

The first (sin(x^3-4x+2))^' is cos(x^3-4x+2)(3x^2-4)

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(sin (x^{3} - 4 x + 2))^{^{'}}

cos (x^{3} - 4 x + 2) (3 x^{2} - 4)

Solution steps

(sin (x^{3} - 4 x + 2))^{'}

Apply the chain rule:

cos (x^{3} - 4 x + 2) (x^{3} - 4 x + 2)^{'}

= cos (x^{3} - 4 x + 2) (x^{3} - 4 x + 2)^{'}

(x^{3} - 4 x + 2)^{'} = 3 x^{2} - 4

= cos (x^{3} - 4 x + 2) (3 x^{2} - 4)

d er i v a t i v e cot^{4} (x)

d er i v a t i v e \frac{1}{( z - 1 )}

d er i v a t i v e y = 25

d er i v a t i v e sin^{35} (x)

\frac{d}{d v} (\frac{( v - 0 )}{t})

What is the (sin(x^3-4x+2))^' ?
The (sin(x^3-4x+2))^' is cos(x^3-4x+2)(3x^2-4)
What is the first (sin(x^3-4x+2))^' ?
The first (sin(x^3-4x+2))^' is cos(x^3-4x+2)(3x^2-4)