What is the derivative of bxsin(x) ?

The derivative of bxsin(x) is b(sin(x)+xcos(x))

What is the first derivative of bxsin(x) ?

The first derivative of bxsin(x) is b(sin(x)+xcos(x))

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\frac{d}{d x} (b x sin (x))

b (sin (x) + x cos (x))

Solution steps

\frac{d}{d x} (b x sin (x))

Take the constant out:

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

= b \frac{d}{d x} (x sin (x))

Apply the Product Rule:

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

f = x, g = sin (x)

= b (\frac{d x}{d x} sin (x) + \frac{d}{d x} (sin (x)) x)

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

\frac{d}{d x} (sin (x)) = cos (x)

= b (1 \cdot sin (x) + cos (x) x)

Simplify

= b (sin (x) + x cos (x))

x^{4} + 9 x

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

x \to 0 + lim (\frac{x ^{4 x}}{7 x})

\int u^{- \frac{2}{3}} d u

\int sin (y) d y

What is the derivative of bxsin(x) ?
The derivative of bxsin(x) is b(sin(x)+xcos(x))
What is the first derivative of bxsin(x) ?
The first derivative of bxsin(x) is b(sin(x)+xcos(x))