What is the (cos(x^2)*2x)^' ?

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

What is the first (cos(x^2)*2x)^' ?

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

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(cos (x^{2}) \cdot 2 x)^{^{'}}

2 (- 2 x^{2} sin (x^{2}) + cos (x^{2}))

Solution steps

(cos (x^{2}) \cdot 2 x)^{'}

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = cos (x^{2}), g = x

= 2 ((cos (x^{2}))^{^{'}} x + x^{^{'}} cos (x^{2}))

(cos (x^{2}))^{'} = - sin (x^{2}) \cdot 2 x

x^{'} = 1

= 2 ((- sin (x^{2}) \cdot 2 x) x + 1 \cdot cos (x^{2}))

Simplify

2 ((- sin (x^{2}) \cdot 2 x) x + 1 \cdot cos (x^{2})) : 2 (- 2 x^{2} sin (x^{2}) + cos (x^{2}))

= 2 (- 2 x^{2} sin (x^{2}) + cos (x^{2}))

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + x y))

\frac{\partial}{\partial y} (2 k y - 5 x)

\int y^{- \frac{1}{3}} d y

\int \frac{5 x - 12}{x ^{3} - 6 x ^{2} + 8 x} d x

\int \frac{x - 3}{x ^{2} + 1} d x

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