What is the derivative of 4x^2cos(x+9x) ?

The derivative of 4x^2cos(x+9x) is 4(2xcos(x)-x^2sin(x))+9

What is the first derivative of 4x^2cos(x+9x) ?

The first derivative of 4x^2cos(x+9x) is 4(2xcos(x)-x^2sin(x))+9

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\frac{d}{d x} (4 x^{2} cos (x) + 9 x)

4 (2 x cos (x) - x^{2} sin (x)) + 9

Solution steps

\frac{d}{d x} (4 x^{2} cos (x) + 9 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{2} cos (x)) + \frac{d}{d x} (9 x)

\frac{d}{d x} (4 x^{2} cos (x)) = 4 (2 x cos (x) - x^{2} sin (x))

\frac{d}{d x} (9 x) = 9

= 4 (2 x cos (x) - x^{2} sin (x)) + 9

\int_{0}^{1} 8 t e^{5 t^{2}} d t

x \to - 2 lim (\frac{3 x - 9}{x + 1})

n = 2 \sum \infty \frac{1}{( - 5 ) ^{n}}

n \to + 0 - lim (5 + \frac{4}{n})

What is the derivative of 4x^2cos(x+9x) ?
The derivative of 4x^2cos(x+9x) is 4(2xcos(x)-x^2sin(x))+9
What is the first derivative of 4x^2cos(x+9x) ?
The first derivative of 4x^2cos(x+9x) is 4(2xcos(x)-x^2sin(x))+9