What is the derivative of 6(sin(x^2)+2x^2cos(x^2)) ?

The derivative of 6(sin(x^2)+2x^2cos(x^2)) is 6(6xcos(x^2)-4x^3sin(x^2))

What is the first derivative of 6(sin(x^2)+2x^2cos(x^2)) ?

The first derivative of 6(sin(x^2)+2x^2cos(x^2)) is 6(6xcos(x^2)-4x^3sin(x^2))

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derivative of

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

6 (6 x cos (x^{2}) - 4 x^{3} sin (x^{2}))

Solution steps

\frac{d}{d x} (6 (sin (x^{2}) + 2 x^{2} cos (x^{2})))

Take the constant out:

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

= 6 \frac{d}{d x} (sin (x^{2}) + 2 x^{2} cos (x^{2}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= 6 (\frac{d}{d x} (sin (x^{2})) + \frac{d}{d x} (2 x^{2} cos (x^{2})))

\frac{d}{d x} (sin (x^{2})) = cos (x^{2}) \cdot 2 x

\frac{d}{d x} (2 x^{2} cos (x^{2})) = 2 (2 x cos (x^{2}) - 2 x^{3} sin (x^{2}))

= 6 (cos (x^{2}) \cdot 2 x + 2 (2 x cos (x^{2}) - 2 x^{3} sin (x^{2})))

Expand

cos (x^{2}) \cdot 2 x + 2 (2 x cos (x^{2}) - 2 x^{3} sin (x^{2})) : 6 x cos (x^{2}) - 4 x^{3} sin (x^{2})

= 6 (6 x cos (x^{2}) - 4 x^{3} sin (x^{2}))

\frac{d}{d x} (\frac{x ^{2} tan ( x ^{3} - x ^{2} )}{sec ( x )})

y^{^{''}} + 3 y^{^{'}} = 10 sin (t)

y = 5 x - 4 x, (1, 1)

\int_{0}^{1} (∣ 3 x - 1 ∣ - 2 ∣ x ∣) d x

- \frac{5}{4} t^{4} + 120 t^{2}

What is the derivative of 6(sin(x^2)+2x^2cos(x^2)) ?
The derivative of 6(sin(x^2)+2x^2cos(x^2)) is 6(6xcos(x^2)-4x^3sin(x^2))
What is the first derivative of 6(sin(x^2)+2x^2cos(x^2)) ?
The first derivative of 6(sin(x^2)+2x^2cos(x^2)) is 6(6xcos(x^2)-4x^3sin(x^2))