What is the first (sin^2(x))^' ?

The first (sin^2(x))^' is sin(2x)

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(sin^2(x))^'

(sin^{2} (x))^{^{'}}

sin (2 x)

Solution steps

(sin^{2} (x))^{'}

Apply the chain rule:

2 sin (x) (sin (x))^{'}

= 2 sin (x) (sin (x))^{'}

(sin (x))^{'} = cos (x)

= 2 sin (x) cos (x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= sin (2 x)

\frac{d}{d x} (4 x^{\frac{1}{3}})

5 x - x^{2}, 0

h \to 0 lim (h)

2 x^{2} y + x^{3} (\frac{d y}{d x}) = 1

f (x) = \frac{ln ( x )}{x}