What is the derivative of (t(cos(t)+8t+2sin(3t))) ?

The derivative of (t(cos(t)+8t+2sin(3t))) is 16x-xsin(x)+6xcos(3x)+cos(x)+2sin(3x)

What is the first derivative of (t(cos(t)+8t+2sin(3t))) ?

The first derivative of (t(cos(t)+8t+2sin(3t))) is 16x-xsin(x)+6xcos(3x)+cos(x)+2sin(3x)

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\frac{d}{d x} ((x) (cos (x) + 8 x + 2 sin (3 x)))

16 x - x sin (x) + 6 x cos (3 x) + cos (x) + 2 sin (3 x)

Solution steps

\frac{d}{d x} ((x) (cos (x) + 8 x + 2 sin (3 x)))

Apply the Product Rule:

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

f = x, g = cos (x) + 8 x + 2 sin (3 x)

= \frac{d x}{d x} (cos (x) + 8 x + 2 sin (3 x)) + \frac{d}{d x} (cos (x) + 8 x + 2 sin (3 x)) x

\frac{d x}{d x} = 1

\frac{d}{d x} (cos (x) + 8 x + 2 sin (3 x)) = - sin (x) + 8 + 6 cos (3 x)

= 1 \cdot (cos (x) + 8 x + 2 sin (3 x)) + (- sin (x) + 8 + 6 cos (3 x)) x

Simplify

1 \cdot (cos (x) + 8 x + 2 sin (3 x)) + (- sin (x) + 8 + 6 cos (3 x)) x : 16 x - x sin (x) + 6 x cos (3 x) + cos (x) + 2 sin (3 x)

= 16 x - x sin (x) + 6 x cos (3 x) + cos (x) + 2 sin (3 x)

\frac{4 x y}{( x ^{2} + 1 ) ( y ^{2} + 1 )}

(1, 3) y (- 1, - 2)

\frac{4 x ^{2}}{( 1 + x ^{2} ) ^{3}}

\frac{2 ( x ^{2} - 1 )}{x ^{2} - 3}

\frac{4 a x}{( x + a ) ^{2} ( x - a ) ^{2}}

What is the derivative of (t(cos(t)+8t+2sin(3t))) ?
The derivative of (t(cos(t)+8t+2sin(3t))) is 16x-xsin(x)+6xcos(3x)+cos(x)+2sin(3x)
What is the first derivative of (t(cos(t)+8t+2sin(3t))) ?
The first derivative of (t(cos(t)+8t+2sin(3t))) is 16x-xsin(x)+6xcos(3x)+cos(x)+2sin(3x)